NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.4

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Chapter 9 Ex.9.4 Question 1

Find the general solution for the differential equation:

\(\frac{{dy}}{{dx}} = \frac{{1 - \cos x}}{{1 + \cos x}}\)

Solution

\[\begin{align}&\frac{{dy}}{{dx}} = \frac{{1 - \cos x}}{{1 + \cos x}}\\& \Rightarrow \frac{{dy}}{{dx}} = \frac{{2{{\sin }^2}\frac{x}{2}}}{{2{{\cos }^2}\frac{x}{2}}} = {\tan ^2}\frac{x}{2}\\& \Rightarrow \frac{{dy}}{{dx}} = \left( {{{\sec }^2}\frac{x}{2} - 1} \right)\\& \Rightarrow \; dy = \left( {{{\sec }^2}\frac{x}{2} - 1} \right)dx\end{align}\]

Integrating both sides, we get:

\[\begin{align}&\int {dy} = \int {\left( {{{\sec }^2}\frac{x}{2} - 1} \right)dx} \\ &\Rightarrow \; y = \int {{{\sec }^2}\frac{x}{2}dx} - \int {dx} \\ &\Rightarrow \; y = 2\tan \frac{x}{2} - x + C\end{align}\]

Chapter 9 Ex.9.4 Question 2

Find the general solution for the differential equation:

\(\frac{{dy}}{{dx}} = \sqrt {4 - {y^2}} \left( { - 2 < y < 2} \right)\)

Solution

\[\begin{align}&\frac{{dy}}{{dx}} = \sqrt {4 - {y^2}} \\ &\Rightarrow \; \frac{{dy}}{{\sqrt {4 - {y^2}} }} = dx\end{align}\]

Integrating both sides, we get:

\[\begin{align}&\int {\frac{{dy}}{{\sqrt {4 - {y^2}} }}} = \int {dx} \\& \Rightarrow \; {\sin ^{ - 1}}\frac{y}{2} = x + C\\ &\Rightarrow \; \frac{y}{2} = \sin \left( {x + C} \right)\\& \Rightarrow \; y = 2\sin \left( {x + C} \right)\end{align}\]

Chapter 9 Ex.9.4 Question 3

Find the general solution for the differential equation:

\(\frac{{dy}}{{dx}} + y = 1\left( {y \ne 1} \right)\)

Solution

\[\begin{align}&\frac{{dy}}{{dx}} + y = 1\left( {y \ne 1} \right)\\ &\Rightarrow \; dy + ydx = dx\\& \Rightarrow \; dy = \left( {1 - y} \right)dx\\& \Rightarrow \; \frac{{dy}}{{1 - y}} = dx\end{align}\]

Integrating both sides, we get:

\[\begin{align}&\int {\frac{{dy}}{{1 - y}}} = \int {dx} \\ &\Rightarrow \; - \log \left( {y - 1} \right) = x + \log C\\ &\Rightarrow \; - \log C - \log \left( {y - 1} \right) = x\\ &\Rightarrow \; - \left[ {\log C + \log \left( {y - 1} \right)} \right] = x\\& \Rightarrow \; \log C\left( {y - 1} \right) = - x\\ &\Rightarrow \; C\left( {y - 1} \right) = {e^{ - x}}\\& \Rightarrow \; y = 1 + \frac{1}{C}{e^{ - x}}\\& \Rightarrow \; y = 1 + A{e^{ - x}}\;\;\;\;\;\;\;\;\;\;\;\left( {{\text{where }}A = \frac{1}{C}} \right)\end{align}\]

Chapter 9 Ex.9.4 Question 4

Find the general solution for the differential equation:

\({\sec ^2}x\tan ydx + {\sec ^2}y\tan xdy = 0\)

Solution

\({\sec ^2}x\tan ydx + {\sec ^2}y\tan xdy = 0\)

Dividing both sides by \(\tan x\tan y\)

\[\begin{align} &\Rightarrow \; \frac{{{{\sec }^2}x\tan ydx + {{\sec }^2}y\tan xdy}}{{\tan x\tan y}} = \frac{0}{{\tan x\tan y}}\\ &\Rightarrow \; \frac{{{{\sec }^2}x}}{{\tan x}}dx + \frac{{{{\sec }^2}y}}{{\tan y}}dy = 0\\ &\Rightarrow \; \frac{{{{\sec }^2}x}}{{\tan x}}dx = - \frac{{{{\sec }^2}y}}{{\tan y}}dy\end{align}\]

Integrating both sides, we get:

\[\int {\frac{{{{\sec }^2}x}}{{\tan x}}dx = - \int {\frac{{{{\sec }^2}y}}{{\tan y}}dy \qquad \ldots \left( {\text{1}} \right)} } \]

Let \(\tan x = t\)

\[\begin{align}& \Rightarrow \; \frac{d}{{dx}}\left( {\tan x} \right) = \frac{{dt}}{{dx}}\\ &\Rightarrow \; {\sec ^2}x = \frac{{dt}}{{dx}}\\& \Rightarrow \; {\sec ^2}xdx = dt\end{align}\]

Now,

\[\begin{align}&\int {\frac{{{{\sec }^2}x}}{{\tan x}}dx} = \int {\frac{1}{t}dt} \\ &= \log t\\& = \log \left( {\tan x} \right) \qquad \ldots \left( 2 \right)\end{align}\]

Similarly,

\[\int {\frac{{{{\sec }^2}y}}{{\tan y}}dy} = \log \left( {\tan y} \right) \qquad \ldots \left( 3 \right)\]

Using \(\left( 1 \right)\), \(\left( 2 \right)\) and \(\left( 3 \right)\)

\[\begin{align}& \Rightarrow \; \log \left( {\tan x} \right) = - \log \left( {\tan y} \right) + \log C\\ &\Rightarrow \; \log \left( {\tan x} \right) = \log \left( {\frac{C}{{\tan y}}} \right)\\ &\Rightarrow \; \tan x = \frac{C}{{\tan y}}\\ &\Rightarrow \; \tan x\tan y = C\end{align}\]

Chapter 9 Ex.9.4 Question 5

Find the general solution for the differential equation:

\(\left( {{e^x} + {e^{ - x}}} \right)dy - \left( {{e^x} - {e^{ - x}}} \right)dx = 0\)

Solution

\[\begin{align}&\left( {{e^x} + {e^{ - x}}} \right)dy - \left( {{e^x} - {e^{ - x}}} \right)dx = 0\\& \Rightarrow \; \left( {{e^x} + {e^{ - x}}} \right)dy = \left( {{e^x} - {e^{ - x}}} \right)dx\\& \Rightarrow \; dy = \left[ {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}} \right]dx\end{align}\]

Integrating both sides, we get:

\[\int {dy = \int {\left[ {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}} \right]dx} } \qquad \ldots \left( 1 \right)\]

Let \(\left( {{e^x} + {e^{ - x}}} \right) = t\)

\[\begin{align}& \Rightarrow \; \frac{d}{{dx}}\left( {{e^x} + {e^{ - x}}} \right) = \frac{{dt}}{{dx}}\\& \Rightarrow \; \left( {{e^x} - {e^{ - x}}} \right)dx = dt\end{align}\]

Putting these values in equation \(\left( 1 \right)\), we get:

\[\begin{align}\int {dy} &= \int {\frac{1}{t}dt + C} \\ &\Rightarrow \; y = \log \left( t \right) + C\\& \Rightarrow \; \log \left( {{e^x} + {e^{ - x}}} \right) + C\end{align}\]

Chapter 9 Ex.9.4 Question 6

Find the general solution for the differential equation:

\(\frac{{dy}}{{dx}} = \left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)\)

Solution

\[\begin{align}\frac{{dy}}{{dx}} &= \left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)\\ &\Rightarrow \; \frac{{dy}}{{1 + {y^2}}} = \left( {1 + {x^2}} \right)dx\end{align}\]

Integrating both sides, we get:

\[\begin{align}&\int {\frac{{dy}}{{1 + {y^2}}} = \int {\left( {1 + {x^2}} \right)dx} } \\ &\Rightarrow \; {\tan ^{ - 1}}y = \int {dx} + \int {{x^2}dx} \\ &\Rightarrow \; {\tan ^{ - 1}}y = x + \frac{{{x^3}}}{x} + C\end{align}\]

Chapter 9 Ex.9.4 Question 7

Find the general solution for the differential equation:

\(y\log ydx - xdy = 0\)

Solution

\[\begin{align}&y\log ydx - xdy = 0\\& \Rightarrow \; y\log ydx = xdy\\& \Rightarrow \; \frac{{dy}}{{y\log y}} = \frac{{dx}}{x}\end{align}\]

Integrating both sides, we get:

\[\int {\frac{{dy}}{{y\log y}} = \int {\frac{{dx}}{x}} }  \qquad \ldots \left( 1 \right)\]

Let \(\log y = t\)

\[\begin{align}& \Rightarrow \; \frac{d}{{dy}}\left( {\log y} \right) = \frac{{dt}}{{dy}}\\ &\Rightarrow \; \frac{1}{y} = \frac{{dt}}{{dy}}\\& \Rightarrow \; \frac{1}{y}dy = dt\end{align}\]

Putting these values in equation \(\left( 1 \right)\), we get:

\[\begin{align}&\int {\frac{{dt}}{t} = \int {\frac{{dx}}{x}} } \\ &\Rightarrow \; \log t = \log x + \log C\\ &\Rightarrow \; \log \left( {\log y} \right) = \log Cx\\& \Rightarrow \; \log y = Cx\\ &\Rightarrow \; y = {e^{Cx}}\end{align}\]

Chapter 9 Ex.9.4 Question 8

Find the general solution for the differential equation:

\({x^5}\frac{{dy}}{{dx}} = - {y^5}\)

Solution

\[\begin{align}&{x^5}\frac{{dy}}{{dx}} = - {y^5}\\& \Rightarrow \; \frac{{dy}}{{{y^5}}} = - \frac{{dx}}{{{x^5}}}\\ &\Rightarrow \; \frac{{dx}}{{{x^5}}} + \frac{{dy}}{{{y^5}}} = 0\end{align}\]

Integrating both sides, we get:

\[\begin{align}&\int {\frac{{dx}}{{{x^5}}} + \int {\frac{{dy}}{{{y^5}}}} } = k\\ &\Rightarrow \; \int {{x^{ - 5}}dx} + \int {{y^{ - 5}}dy} = k\\ &\Rightarrow \; \frac{{{x^{ - 4}}}}{{ - 4}} + \frac{{{y^{ - 4}}}}{{ - 4}} = k\\ &\Rightarrow \; {x^{ - 4}} + {y^{ - 4}} = - 4k\\& \Rightarrow \; {x^{ - 4}} + {y^{ - 4}} = C{\text{ }}\left( {{\text{where }}C = - 4k} \right)\end{align}\]

Chapter 9 Ex.9.4 Question 9

Find the general solution for the differential equation:

\(\frac{{dy}}{{dx}} = {\sin ^{ - 1}}x\)

Solution

\[\begin{align}&\frac{{dy}}{{dx}} = {\sin ^{ - 1}}x\\ &\Rightarrow \; dy = {\sin ^{ - 1}}xdx\end{align}\]

Integrating both sides, we get:

\[\begin{align}\int {dy}& = \int {{{\sin }^{ - 1}}xdx} \\ &\Rightarrow \; y = \int {\left( {{{\sin }^{ - 1}}x.1} \right)dx} \\ &\Rightarrow \; y = {\sin ^{ - 1}}x.\int {\left( 1 \right)dx} - \int {\left[ {\left( {\frac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right)\int {\left( 1 \right)dx} } \right)} \right]dx} \\ &\Rightarrow \; y = x{\sin ^{ - 1}}x + \int {\frac{{ - x}}{{\sqrt {1 - {x^2}} }}dx} \qquad \ldots \left( 1 \right)\end{align}\]

Let \(1 - {x^2} = t\)

\[\begin{align}& \Rightarrow \; \frac{d}{{dx}}\left( {1 - {x^2}} \right) = \frac{{dt}}{{dx}}\\& \Rightarrow \; - 2x = \frac{{dt}}{{dx}}\\ &\Rightarrow \; xdx = - \frac{1}{2}dt\end{align}\]

Putting these values in equation \(\left( 1 \right)\), we get:

\[\begin{align} &\Rightarrow \; y = x{\sin ^{ - 1}}x + \int {\frac{1}{{2\sqrt t }}} dt\\ &\Rightarrow \; y = x{\sin ^{ - 1}}x + \frac{1}{2}{\int {\left( t \right)} ^{\frac{{ - 1}}{2}}}dt\\& \Rightarrow \; y = x{\sin ^{ - 1}}x + \frac{1}{2}.\left( {\frac{{{t^{\frac{1}{2}}}}}{{\frac{1}{2}}}} \right) + C\\ &\Rightarrow \; y = x{\sin ^{ - 1}}x + \sqrt t + C\\&y = x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} + C\end{align}\]

Chapter 9 Ex.9.4 Question 10

Find the general solution for the differential equation:

\({e^x}\tan ydx + \left( {1 - {e^x}} \right){\sec ^2}ydy = 0\)

Solution

\[\begin{align}&{e^x}\tan ydx + \left( {1 - {e^x}} \right){\sec ^2}ydy = 0\\&\left( {1 - {e^x}} \right){\sec ^2}ydy = - {e^x}\tan ydx\\&\frac{{{{\sec }^2}y}}{{\tan y}}dy = \frac{{ - {e^x}}}{{1 - {e^x}}}dx\end{align}\]

Integrating both sides, we get:

\[\int {\frac{{{{\sec }^2}y}}{{\tan y}}dy} = \int {\frac{{ - {e^x}}}{{1 - {e^x}}}dx} \qquad \ldots \left( 1 \right)\]

Let \(\tan y = u\)

\[\begin{align} &\Rightarrow \; \frac{d}{{dy}}\left( {\tan y} \right) = \frac{{du}}{{dy}}\\ &\Rightarrow \; {\sec ^2}y = \frac{{du}}{{dy}}\\& \Rightarrow \; {\sec ^2}ydy = du\end{align}\]

Integrating both sides, we get:

\[\begin{align}\int {\frac{{{{\sec }^2}y}}{{\tan y}}dy}& = \int {\frac{{du}}{u}} \\ &= \log u\\ &= \log \left( {\tan y} \right) \qquad \ldots \left( 2 \right)\end{align}\]

Now, let \(\left( {1 - {e^x}} \right) = t\)

\[\begin{align}&\frac{d}{{dx}}\left( {1 - {e^x}} \right) = \frac{{dt}}{{dx}}\\& \Rightarrow \; - {e^x} = \frac{{dt}}{{dx}}\\& \Rightarrow \; - {e^x}dx = dt\end{align}\]

Integrating both sides, we get:

\[\begin{align}\int {\frac{{ - {e^x}}}{{1 - {e^x}}}} dx &= \int {\frac{{dt}}{t}} \\& = \log t\\& = \log \left( {1 - {e^x}} \right) \qquad \ldots \left( 3 \right)\end{align}\]

Using \(\left( 1 \right)\), \(\left( 2 \right)\) and \(\left( 3 \right)\)

\[\begin{align} &\Rightarrow \; \log \left( {\tan y} \right) = \log \left( {1 - {e^x}} \right) + \log C\\ &\Rightarrow \; \log \left( {\tan y} \right) = \log \left[ {C\left( {1 - {e^x}} \right)} \right]\\& \Rightarrow \; \tan y = C\left( {1 - {e^x}} \right)\end{align}\]

Chapter 9 Ex.9.4 Question 11

\(\left( {{x^3} + {x^2} + x + 1} \right)\frac{{dy}}{{dx}} = 2{x^2} + x\;;y = 1\) when \(x = 0\)

Solution

\[\begin{align}&\left( {{x^3} + {x^2} + x + 1} \right)\frac{{dy}}{{dx}} = 2{x^2} + x;y = 1\\& \Rightarrow \; \frac{{dy}}{{dx}} = \frac{{2{x^2} + x}}{{\left( {{x^3} + {x^2} + x + 1} \right)}}\\ &\Rightarrow \; dy = \frac{{2{x^2} + x}}{{\left( {{x^3} + {x^2} + x + 1} \right)}}dx\end{align}\]

Integrating both sides, we get:

\[\int {dy} = \int {\frac{{2{x^2} + x}}{{\left( {{x^3} + {x^2} + x + 1} \right)}}dx} \qquad \ldots \left( 1 \right)\]

Let \(\frac{{2{x^2} + x}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} = \frac{A}{{x + 1}} + \frac{{Bx + C}}{{{x^2} + 1}} \qquad \ldots \left( 2 \right)\)

\[\begin{align} &\Rightarrow \; \frac{{2{x^2} + x}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} = \frac{{A{x^2} + A + \left( {Bx + C} \right)\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}}\\ &\Rightarrow \; 2{x^2} + x = A{x^2} + A + B{x^2} + Bx + Cx + C\\ &\Rightarrow \; 2{x^2} + x = \left( {A + B} \right){x^2} + \left( {B + C} \right)x + \left( {A + C} \right)\end{align}\]

Comparing the coefficients of \({x^2}\) and \(x\), we get:

\[\begin{align}A + B &= 2\\B + C &= 2\\A + C &= 0\end{align}\]

Therefore,

\(A = \frac{1}{2},B = \frac{3}{2}\) and \(C = \frac{{ - 1}}{2}\)

Substituting these values in \(\left( 2 \right)\), we get:

\[\frac{{2{x^2} + x}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} = \frac{1}{2}\frac{1}{{\left( {x + 1} \right)}} + \frac{1}{2}\frac{{\left( {3x - 1} \right)}}{{\left( {{x^2} + 1} \right)}}\]

Hence, equation \(\left( 1 \right)\)becomes:

\[\begin{align}\int {dy} &= \frac{1}{2}\int {\frac{1}{{\left( {x + 1} \right)}}dx} + \frac{1}{2}\int {\frac{{\left( {3x - 1} \right)}}{{\left( {{x^2} + 1} \right)}}dx} \\y &= \frac{1}{2}\log \left( {x + 1} \right) + \frac{3}{2}\int {\frac{x}{{{x^2} + 1}}} dx - \frac{1}{2}\int {\frac{x}{{{x^2} + 1}}} dx\\ &= \frac{1}{2}\log \left( {x + 1} \right) + \frac{3}{4}\int {\frac{{2x}}{{{x^2} + 1}}} dx - \frac{1}{2}{\tan ^{ - 1}}x + C\\ &= \frac{1}{4}\left[ {2\log \left( {x + 1} \right) + 3\log \left( {{x^2} + 1} \right)} \right] - \frac{1}{2}{\tan ^{ - 1}}x + C\\ &= \frac{1}{4}\left[ {\log {{\left( {x + 1} \right)}^2} + \log {{\left( {{x^2} + 1} \right)}^3}} \right] - \frac{1}{2}{\tan ^{ - 1}}x + C\\ &= \frac{1}{4}\left[ {\log {{\left( {x + 1} \right)}^2}{{\left( {{x^2} + 1} \right)}^3}} \right] - \frac{1}{2}{\tan ^{ - 1}}x + C\end{align}\]

Now, \(y = 1\) when \(x = 0\)

\[\begin{align}& \Rightarrow \; 1 = \frac{1}{4}\log \left( 1 \right) - \frac{1}{2}{\tan ^{ - 1}}0 + C\\ &\Rightarrow \; 1 = \frac{1}{4} \times 0 - \frac{1}{2} \times 0 + C\\ &\Rightarrow \; C = 1\end{align}\]

Thus, \(y = \frac{1}{4}\left[ {\log {{\left( {x + 1} \right)}^2}{{\left( {{x^2} + 1} \right)}^3}} \right] - \frac{1}{2}{\tan ^{ - 1}}x + 1\)

Chapter 9 Ex.9.4 Question 12

For the below differential equation find a particular solution satisfying the given condition:

\(x\left( {{x^2} - 1} \right)\frac{{dy}}{{dx}} = 1;\;y = 0\) when \(x = 2\)

Solution

\[\begin{align}&x\left( {{x^2} - 1} \right)\frac{{dy}}{{dx}} = 1\\& \Rightarrow \; dy = \frac{{dx}}{{x\left( {{x^2} - 1} \right)}}\\ &\Rightarrow \; dy = \frac{1}{{x\left( {x - 1} \right)\left( {x + 1} \right)}}dx\end{align}\]

Integrating both sides, we get:

\[\int {dy} = \int {\frac{1}{{x\left( {x - 1} \right)\left( {x + 1} \right)}}dx} \qquad \ldots \left( 1 \right)\]

Let \(\frac{1}{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{A}{x} + \frac{B}{{x - 1}} + \frac{C}{{x + 1}} \qquad \ldots \left( 2 \right)\)

\[\begin{align} &\Rightarrow \; \frac{1}{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{{A\left( {x - 1} \right)\left( {x + 1} \right) + Bx\left( {x + 1} \right) + Cx\left( {x - 1} \right)}}{{x\left( {x - 1} \right)\left( {x + 1} \right)}}\\& \Rightarrow \; \frac{1}{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{{\left( {A + B + C} \right){x^2} + \left( {B - C} \right)x - A}}{{x\left( {x - 1} \right)\left( {x + 1} \right)}}\end{align}\]

Comparing the coefficients of \({x^2}\) and \(x\), we get:

\[\begin{align}&A = - 1\\&B - C = 0\\&A + B + C = 0\end{align}\]

Therefore,

\(A = - 1,\;B = \frac{1}{2}\) and \(C = \frac{1}{2}\)

Substituting these values in \(\left( 2 \right)\), we get:

\( \Rightarrow \; \frac{1}{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{{ - 1}}{x} + \frac{1}{{2\left( {x - 1} \right)}} + \frac{1}{{2\left( {x + 1} \right)}}\)

Hence, equation \(\left( 1 \right)\)becomes:

\[\begin{align}&\int {dy} = - \int {\frac{1}{x}dx} + \frac{1}{2}\int {\frac{1}{{x - 1}}dx} + \frac{1}{2}\int {\frac{1}{{x + 1}}dx} \\ &\Rightarrow \; y = - \log x + \frac{1}{2}\log \left( {x - 1} \right) + \frac{1}{2}\log \left( {x + 1} \right) + \log k\\& \Rightarrow \; y = \frac{1}{2}\log \left[ {\frac{{{k^2}\left( {x - 1} \right)\left( {x + 1} \right)}}{{{x^2}}}} \right]\end{align}\]

Now, \(y = 0\) when \(x = 2\)

\[\begin{align}&\Rightarrow \; 0 = \frac{1}{2}\log \left[ {\frac{{{k^2}\left( {2 - 1} \right)\left( {2 + 1} \right)}}{4}} \right]\\& \Rightarrow \; \log \left( {\frac{{3{k^2}}}{4}} \right) = 0\\& \Rightarrow \; \frac{{3{k^2}}}{4} = 1\\ &\Rightarrow \; 3{k^2} = 4\\ &\Rightarrow \; {k^2} = \frac{4}{3}\end{align}\]

Thus,

\[\begin{align}y &= \frac{1}{2}\log \left[ {\frac{{4\left( {x - 1} \right)\left( {x + 1} \right)}}{{3{x^2}}}} \right]\\ &= \frac{1}{2}\log \left[ {\frac{{4\left( {{x^2} - 1} \right)}}{{3{x^2}}}} \right]\\ &= \frac{1}{2}\log \left[ {\frac{4}{3}.\frac{{\left( {{x^2} - 1} \right)}}{{{x^2}}}} \right]\\ &= \frac{1}{2}\log \left( {\frac{{{x^2} - 1}}{{{x^2}}}} \right) - \frac{1}{2}\log \frac{3}{4}\end{align}\]

Chapter 9 Ex.9.4 Question 13

For the below differential equation find a particular solution satisfying the given condition:

\(\cos \left( {\frac{{dy}}{{dx}}} \right) = a\left( {a \in R} \right);\;y = 1\) when \(x = 0\)

Solution

\[\begin{align}&\cos \left( {\frac{{dy}}{{dx}}} \right) = a\\ &\Rightarrow \; \frac{{dy}}{{dx}} = {\cos ^{ - 1}}a\\ &\Rightarrow \; dy = {\cos ^{ - 1}}adx\end{align}\]

Integrating both sides, we get:

\[\begin{align}&\int {dy = {{\cos }^{ - 1}}a\int {dx} } \\& \Rightarrow \; y = {\cos ^{ - 1}}a.x + C\\ &\Rightarrow \; y = x{\cos ^{ - 1}}a + C\end{align}\]

Now, \(y = 1\) when \(x = 0\)

\[\begin{align}& \Rightarrow \; 1 = 0.{\cos ^{ - 1}}a + C\\& \Rightarrow \; C = 1\end{align}\]

Thus,

\[\begin{align}y& = x{\cos ^{ - 1}}a + 1\\ &\Rightarrow \; \frac{{y - 1}}{x} = {\cos ^{ - 1}}a\\& \Rightarrow \; \cos \left( {\frac{{y - 1}}{x}} \right) = a\end{align}\]

Chapter 9 Ex.9.4 Question 14

For the below differential equation find a particular solution satisfying the given condition:

\(\frac{{dy}}{{dx}} = y\tan x;\;y = 1\) when \(x = 0\)

Solution

\[\begin{align}\frac{{dy}}{{dx}} &= y\tan x\\& \Rightarrow \; \frac{{dy}}{y} = \tan xdx\end{align}\]

Integrating both sides, we get:

\[\begin{align}&\int {\frac{{dy}}{y} = \int {\tan xdx} } \\& \Rightarrow \; \log y = \log \left( {\sec x} \right) + \log C\\ &\Rightarrow \; \log y = \log \left( {\sec x.C} \right)\\ &\Rightarrow \; y = C\sec x\end{align}\]

Now, \(y = 1\) when \(x = 0\)

\[\begin{align} &\Rightarrow \; 1 = C \times \sec 0\\ &\Rightarrow \; 1 = C \times 1\\& \Rightarrow \; C = 1\end{align}\]

Thus, \(y = \sec x\)

Chapter 9 Ex.9.4 Question 15

Find the equation of a curve passing through the point \(\left( {{\text{0,0}}} \right)\) and whose differential equation is \(y' = {e^x}\sin x\).

Solution

\[\begin{align}y' &= {e^x}\sin x\\&\Rightarrow \; \frac{{dy}}{{dx}} = {e^x}\sin x\\ &\Rightarrow \; dy = {e^x}\sin xdx\end{align}\]

Integrating both sides, we get:

\[\begin{align}\int {dy} &= \int {{e^x}\sin xdx} \\y &= \int {{e^x}\sin xdx} \;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}\]

Let \(I = \int {{e^x}\sin xdx} \)

\[\begin{align}& \Rightarrow \; I = \sin x\int {{e^x}} dx - \int {\left[ {\frac{d}{{dx}}\left( {\sin x} \right).\int {{e^x}dx} } \right]dx} \\ &\Rightarrow \; I = \sin x.{e^x} - \int {\cos x.{e^x}} dx\\& \Rightarrow \; I = \sin x.{e^x} - \left[ {\cos x\int {{e^x}dx} - \int {\left( {\frac{d}{{dx}}\left( {\cos x} \right).\int {{e^x}dx} } \right)dx} } \right]\\& \Rightarrow \; I = \sin x.{e^x} - \left[ {\cos x{e^x} - \int {\left( { - \sin x} \right){e^x}dx} } \right]\\& \Rightarrow \; I = {e^x}\sin x - {e^x}\cos x - I\\ &\Rightarrow \; 2I = {e^x}\left( {\sin x - \cos x} \right)\\ &\Rightarrow \; I = \frac{{{e^x}\left( {\sin x - \cos x} \right)}}{2}\end{align}\]

Substituting this value in equation \(\left( 1 \right)\), we get:

\[y = \frac{{{e^x}\left( {\sin x - \cos x} \right)}}{2} + C\]

Since, the curve passes through \(\left( {0,0} \right)\), we have:

\[\begin{align} &\Rightarrow \; 0 = \frac{{{e^0}\left( {\sin 0 - \cos 0} \right)}}{2} + C\\&\Rightarrow \; 0 = \frac{{1\left( {0 - 1} \right)}}{2} + C\\ &\Rightarrow \; C = \frac{1}{2}\end{align}\]

Thus,

\[\begin{align} &\Rightarrow \; y = \frac{{{e^x}\left( {\sin x - \cos x} \right)}}{2} + \frac{1}{2}\\& \Rightarrow \; 2y = {e^x}\left( {\sin x - \cos x} \right) + 1\\ &\Rightarrow \; 2y - 1 = {e^x}\left( {\sin x - \cos x} \right)\\ &\Rightarrow \; 2y - 1 = {e^x}\left( {\sin x - \cos x} \right)\end{align}\]

Hence, the required equation of the curve is \(2y - 1 = {e^x}\left( {\sin x - \cos x} \right)\)

Chapter 9 Ex.9.4 Question 16

For the differential equation \(xy\frac{{dy}}{{dx}} = \left( {x + 2} \right)\left( {y + 2} \right)\). Find the solution curve passing through the point \(\left( {1, - 1} \right)\).

Solution

\[\begin{align}xy\frac{{dy}}{{dx}}& = \left( {x + 2} \right)\left( {y + 2} \right)\\& \Rightarrow \; \left( {\frac{y}{{y + 2}}} \right)dy = \left( {\frac{{x + 2}}{2}} \right)dx\end{align}\]

Integrating both sides, we get:

\[\begin{align}&\int {\left( {1 - \frac{2}{{y + 2}}} \right)dy = \int {\left( {1 + \frac{2}{x}} \right)} dx} \\ &\Rightarrow \; \int {dy} - 2\int {\frac{1}{{y + 2}}} dy = \int {dx} + 2\int {\frac{1}{x}} dx\\& \Rightarrow \; y - 2\log \left( {y + 2} \right) = x + 2\log x + C\\& \Rightarrow \; y - x - C = \log {x^2} + \log {\left( {y + 2} \right)^2}\\ &\Rightarrow \; y - x - C = \log \left[ {{x^2}{{\left( {y + 2} \right)}^2}} \right]\end{align}\]

Since, the curve passes through \(\left( {1, - 1} \right)\), we have:

\[\begin{align}& \Rightarrow \; 1 - 1 - C = \log \left[ {{{\left( 1 \right)}^2}{{\left( { - 1 + 2} \right)}^2}} \right]\\ &\Rightarrow \; - 2 - C = \log 1\\& \Rightarrow \; - 2 - C = 0\\& \Rightarrow \; C = - 2\end{align}\]

Thus, \(y - x + 2 = \log \left[ {{x^2}{{\left( {y + 2} \right)}^2}} \right]\) is the required solution of the curve.

Chapter 9 Ex.9.4 Question 17

Find the equation of a curve passing through the point \(\left( {0, - 2} \right)\) given that at any point\(\left( {x,y} \right)\) on the curve, the product of the slope of its tangent and \(y\)-coordinate of the point is equal to the \(x\)-coordinate of the point.

Solution

Let \(x\) and \(y\) be \(x\)-coordinate and y-coordinate of the curve, respectively.

We know that the slope of a tangent to the curve in the coordinate axis is given by the relation, \(\frac{{dy}}{{dx}}\)

Therefore

\[\begin{align} &\Rightarrow \; y.\frac{{dy}}{{dx}} = x\\ &\Rightarrow \; ydy = xdx\end{align}\]

Integrating both sides, we get:

\[\begin{align}&\int {ydy = \int {xdx} } \\ &\Rightarrow \; \frac{{{y^2}}}{2} = \frac{{{x^2}}}{2} + C\\ &\Rightarrow \; {y^2} - {x^2} = 2C\end{align}\]

Since, the curve passes through \(\left( {0, - 2} \right)\), we have:

\[\begin{align} &\Rightarrow \; {\left( { - 2} \right)^2} - {0^2} = 2C\\ &\Rightarrow \; 2C = 4\end{align}\]

Thus, \({y^2} - {x^2} = 4\) is the required equation of the curve.

Chapter 9 Ex.9.4 Question 18

At any point \(\left( {x,y} \right)\) of a curve, the slope of the tangent is twice the slope of the line segment joining ate point of contact to the point\(\left( { - 4, - 3} \right)\). Find the equation of the curve given that it passes through \(\left( { - 2,1} \right)\).

Solution

\(\left( {x,y} \right)\) is point of contact of curve and tangent.

Slope \(\left( {{m_1}} \right)\) of segment joining \(\left( {x,y} \right)\) and \(\left( { - 4, - 3} \right)\) is \(\frac{{y + 3}}{{x + 4}}\)

We know that the slope of a tangent to the curve in the coordinate axis is given by the relation, \(\frac{{dy}}{{dx}}\).

Therefore, slope \(\left( {{m_2}} \right)\) of tangent is \(\frac{{dy}}{{dx}}\).

Since, \({m_2} = 2{m_1}\)

\[\begin{align}& \Rightarrow \; \frac{{dy}}{{dx}} = \frac{{2\left( {y + 3} \right)}}{{x + 4}}\\ &\Rightarrow \; \frac{{dy}}{{y + 3}} = \frac{{2dx}}{{x + 4}}\end{align}\]

Integrating both sides, we get:

\[\begin{align}&\int {\frac{{dy}}{{y + 3}} = 2\int {\frac{{dx}}{{x + 4}}} } \\ &\Rightarrow \; \log \left( {y + 3} \right) = 2\log \left( {x + 4} \right) + \log C\\& \Rightarrow \; \log \left( {y + 3} \right) = \log C{\left( {x + 4} \right)^2}\\ &\Rightarrow \; y + 3 = C{\left( {x + 4} \right)^2}\end{align}\]

Since, the curve passes through \(\left( { - 2,1} \right)\), we have:

\[\begin{align}& \Rightarrow \; 1 + 3 = C{\left( { - 2 + 4} \right)^2}\\ &\Rightarrow \; 4 = 4C\\ &\Rightarrow \; C = 1\end{align}\]

Thus, \(y + 3 = {\left( {x + 4} \right)^2}\) is the required equation of the curve.

Chapter 9 Ex.9.4 Question 19

The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is \(3\) units and after \(3\) seconds, it is \(6\) units. Find the radius of balloon after \(t\) seconds.

Solution

Let the rate of change of volume of the balloon be \(k.\)

\[\begin{align} &\Rightarrow \; \frac{{dV}}{{dt}} = k\\ &\Rightarrow \; \frac{d}{{dt}}\left( {\frac{4}{3}\pi {r^3}} \right) = k\\ &\Rightarrow \; \frac{4}{3}\pi .3{r^2}.\frac{{dr}}{{dt}} = k\\& \Rightarrow \; 4\pi {r^2}dr = kdt\end{align}\]

Integrating both sides, we get:

\[\begin{align}&\int {4\pi {r^2}dr = \int {kdt} } \\ &\Rightarrow \; 4\pi \frac{{{r^3}}}{3} = kt + C\\ &\Rightarrow \; 4\pi {r^3} = 3\left( {kt + C} \right)\end{align}\]

At \(t = 0,\;r = 3\)

\[\begin{align} &\Rightarrow \; 4\pi \times 27 = 3\left( {k \times 0 + C} \right)\\ &\Rightarrow \; 108\pi = 3C\\ &\Rightarrow \; C = 36\pi\end{align}\]

Now, at \(t = 3,\;r = 6\)

\[\begin{align} &\Rightarrow \; 4\pi \times {6^3} = 3\left( {k \times 3 + C} \right)\\& \Rightarrow \; 864\pi = 3\left( {3k + 36\pi } \right)\\& \Rightarrow \; 3k = 288\pi - 36\pi = 252\pi \\& \Rightarrow \; k = 84\pi\end{align}\]

Hence,

\[\begin{align}4\pi {r^3} &= 3\left[ {84\pi t + 36\pi } \right]\\4\pi {r^3} &= 4\pi \left( {63t + 27} \right)\\{r^3} &= 63t + 27\\r &= {\left( {63t + 27} \right)^{\frac{1}{3}}}\end{align}\]

Thus, the radius of the balloon after \(t\) seconds is \({\left( {63t + 27} \right)^{\frac{1}{3}}}\).

Chapter 9 Ex.9.4 Question 20

In a bank, principle increases continuously at the rate of \(r\% \) per year. Find the value of \(r\) if ₹ \(100\) doubles itself in \(10\) years.\(\left( {{{\log }_e}2 = 0.6931} \right)\)

Solution

Let \(p,t\) and \(r\) represent the principle, time and rate of interest respectively.

The principle increases continuously at the rate of \(r\% \) per year.

\[\begin{align} &\Rightarrow \; \frac{{dp}}{{dt}} = \left( {\frac{r}{{100}}} \right)p\\& \Rightarrow \; \frac{{dp}}{p} = \left( {\frac{r}{{100}}} \right)dt\end{align}\]

Integrating both sides, we get:

\[\begin{align}&\int {\frac{{dp}}{p} = \frac{r}{{100}}\int {dt} } \\& \Rightarrow \; \log p = \frac{{rt}}{{100}} + k\\& \Rightarrow \; p = {e^{\frac{{rt}}{{100}} + k}}\end{align}\]

It is given that \(p = 100\) when \(t = 0\)

Therefore,

\( \Rightarrow \; 100 = {e^k}\)

Now, if \(t = 10\) then \(p = 2 \times 100 = 200\)

Hence,

\[\begin{align}&200 = {e^{\frac{r}{{10}} + k}}\\& \Rightarrow \; 200 = {e^{\frac{r}{{10}}}}.{e^k}\\ &\Rightarrow \; 200 = {e^{\frac{r}{{10}}}}.100\\& \Rightarrow \; {e^{\frac{r}{{10}}}} = 2\\& \Rightarrow \; \frac{r}{{10}} = {\log _e}2\\ &\Rightarrow \; \frac{r}{{10}} = 0.6931\\& \Rightarrow \; r = 6.931\end{align}\]

Thus, the rate of interest, \(r = 6.931\% \).

Chapter 9 Ex.9.4 Question 21

In a bank, principle increases continuously at the rate of \(5\% \) per year. An amount of ₹ \(1000\) is deposited with this bank, how much will it worth after \(10\) years. \(\left( {{e^{0.5}} = 1.648} \right)\)

Solution

Let \(p\) and \(t\) be the principle and time, respectively.

The principle increases continuously at the rate of \(5\% \) per year.

\[\begin{align} &\Rightarrow \; \frac{{dp}}{{dt}} = 5\% \times p\\& \Rightarrow \; \frac{{dp}}{{dt}} = \left( {\frac{5}{{100}}} \right)p\\ &\Rightarrow \; \frac{{dp}}{{dt}} = \frac{p}{{20}}\\ &\Rightarrow \; \frac{{dp}}{p} = \frac{{dt}}{{20}}\end{align}\]

Integrating both sides, we get:

\[\begin{align}&\int {\frac{{dp}}{p} = \frac{1}{{20}}} \int {dt} \\& \Rightarrow \; \log p = \frac{t}{{20}} + C\\ &\Rightarrow \; p = {e^{\frac{t}{{20}} + C}}\end{align}\]

Now, \(p = 1000\) when \(t = 0\)

Therefore,

\( \Rightarrow \; 1000 = {e^C}\)

Now, at \(t = 10\) and \({e^C} = 1000\)

\[\begin{align} &\Rightarrow \; p = {e^{\frac{{10}}{{20}} + C}}\\ &\Rightarrow \; p = {e^{0.5}} \times {e^C}\\& \Rightarrow \; p = 1.648 \times 1000\\ &\Rightarrow \; p = 1648\end{align}\]

Thus, after \(10\) years the amount will worth ₹ \(1648.\)

Chapter 9 Ex.9.4 Question 22

In a culture, the bacteria count is \(1,00,000\). The number is increased by \(10\%\) in \(2\) hours. In how many hours will the count reach \(2,00,000\), if the rate of growth of bacteria is proportional to the number present?

Solution

Let \(y\) be the number of bacteria at any instant \(t.\)

Rate of growth of the bacteria is proportional to the number present.

\[\begin{align}& \Rightarrow \; \frac{{dy}}{{dt}} \propto y\\& \Rightarrow \; \frac{{dy}}{{dt}} = ky\\ &\Rightarrow \; \frac{{dy}}{y} = kdt\end{align}\]

Integrating both sides, we get:

\[\begin{align}\int {\frac{{dy}}{y}} &= k\int {dt} \\ \Rightarrow \; \log y& = kt + C\end{align}\]

Let \({y_0}\) be the number of bacteria at \(t = 0\).

\[\begin{align} &\Rightarrow \; \log {y_0} = C\\& \Rightarrow \; \log y = kt + \log {y_0}\\& \Rightarrow \; \log y - \log {y_0} = kt\\ &\Rightarrow \; \log \left( {\frac{y}{{{y_0}}}} \right) = kt\\ &\Rightarrow \; kt = \log \left( {\frac{y}{{{y_0}}}} \right)\end{align}\]

Since, the number of bacteria increases by \(10\%\) in \(2\) hours.

\[\begin{align} &\Rightarrow \; y = \frac{{110}}{{100}}{y_0}\\& \Rightarrow \; \frac{y}{{{y_0}}} = \frac{{11}}{{10}}\end{align}\]

Taking log on both the sides

\[\begin{align}&\Rightarrow \; \log \left( {\frac{y}{{{y_0}}}} \right) = \log \left( {\frac{{11}}{{10}}} \right) \hfill \\&\Rightarrow \; kt = \log \left( {\frac{{11}}{{10}}} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {\because \log \left( {\frac{y}{{{y_0}}}} \right) = kt} \right] \hfill \\&\Rightarrow \; k.2 = \log \left( {\frac{{11}}{{10}}} \right) \hfill \\&\Rightarrow \; k = \frac{1}{2}\log \left( {\frac{{11}}{{10}}} \right) \hfill \\\end{align} \]

Therefore,

\[\begin{align} &\Rightarrow \; \frac{1}{2}\log \left( {\frac{{11}}{{10}}} \right)t = \log \left( {\frac{y}{{{y_0}}}} \right)\\ &\Rightarrow \; t = \frac{{2\log \left( {\frac{y}{{{y_0}}}} \right)}}{{\log \left( {\frac{{11}}{{10}}} \right)}}\end{align}\]

Now, let the time when the number of bacteria increases from \(1,00,000\) to \(2,00,000\) be \({t_1}\)

Therefore, \(y = {y_0}\) at \(t = {t_1}\)

Hence,

\[\begin{align}{t_1} &= \frac{{2\log \left( {\frac{y}{{{y_0}}}} \right)}}{{\log \left( {\frac{{11}}{{10}}} \right)}}\\ &= \frac{{2\log \left( {\frac{{200000}}{{100000}}} \right)}}{{\log \left( {\frac{{11}}{{10}}} \right)}}\\ &= \frac{{2\log 2}}{{\log \left( {\frac{{11}}{{10}}} \right)}}\end{align}\]

Thus, in \(\frac{{2\log 2}}{{\log \left( {\frac{{11}}{{10}}} \right)}}\) hours, the number of bacteria increases from \(1,00,000\)  to \(2,00,000.\)

Chapter 9 Ex.9.4 Question 23

The general solution of the differential equation \(\frac{{dy}}{{dx}} = {e^{x + y}}\) is

(A) \({e^x} + {e^{ - y}} = C\)

(B) \({e^x} + {e^y} = C\)

(C) \({e^{ - x}} + {e^y} = C\)

(D) \({e^{ - x}} + {e^{ - y}} = C\)

Solution

\[\begin{align}\frac{{dy}}{{dx}} &= {e^{x + y}} = {e^x}.{e^y}\\& \Rightarrow \; \frac{{dy}}{{{e^y}}} = {e^x}dx\\ &\Rightarrow \; {e^{ - y}}dy = {e^x}dx\end{align}\]

Integrating both sides, we get:

\[\begin{align}&\int {{e^{ - y}}dy = \int {{e^x}dx} } \\& \Rightarrow \; - {e^{ - y}} = {e^x} + k\\ &\Rightarrow \; {e^x} + {e^{ - y}} = - k\\& \Rightarrow \; {e^x} + {e^{ - y}} = C\left( {\text{ where, }}{{\text{ C}} = - k} \right)\end{align}\]

Thus, the correct option is (A).

  
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