# NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.4

Go back to  'Differential Equations'

## 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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