Differential Equations Set-1

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As described in the introduction, differential equations are so important for the very reason that they find a wide application in studying all sorts of scientific phenomena. The motion of a body in a force field, radioactive decay and population growth were some of the phenomena mentioned that must use DEs for analysis.

In some of the subsequent solved examples, we apply the DE- solving techniques that we’ve learnt in the previous section, to solve practical and interesting problems.

Example – 1

Let \(f:{\mathbb{R}^ + } \to \mathbb{R}\) be a differentiable function such that

\[f\left( x \right) = e - \left( {x - 1} \right)\ln \frac{x}{e} + \int\limits_1^x {f\left( x \right)dx}\]

Find a simple expression for f(x).

Solution: Differentiating the given relation, we have

\[\begin{gathered} \qquad \frac{{df}}{{dx}} = - \frac{{\left( {x - 1} \right)}}{x} - \ln \frac{x}{e} + f \\ \Rightarrow \quad \frac{{df}}{{dx}} - f = \frac{1}{x} - \ln x \\ \end{gathered}\]

This is evidently a first-order linear DE; the IF is \({e^{\int { - dx} }} = {e^{ - x}}.\) Multiplying it across both sides of the DE renders the DE exact and its solution is given by

\[\begin{align}  {e^{ - x}} \cdot f &= \int {{e^{ - x}}\left( {\frac{1}{x} - \ln x} \right)dx}\\   &= {e^{ - x}}\ln x + C\\   \Rightarrow\qquad f\left( x \right) &= \,\ln x + C\,{e^x} &  \ldots (1)\\ \end{align} \]

From the relation specified in the equation, note that

\[\begin{align}  f\left( 1 \right)&= e - \left( {1 - 1} \right)\left( {\ln \frac{1}{e}} \right) + \int\limits_1^1 {f\left( x \right)dx} \\ &= e\end{align} \]

From (1), \(f(1) = Ce.\) This gives C = 1.

Thus, the function f(x) has the simple form

\[f\left( x \right) = \ln x + {e^x}\]

Example – 2

Solve the following DEs 

(a) \(\begin{align}\frac{{dy}}{{dx}} = \frac{{\cos x\left( {2\cos y - {{\sin }^2}x} \right)}}{{\sin y}}\end{align}\)

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

Solution:(a) We have,

\[\sin y\frac{{dy}}{{dx}} - 2\cos y\cos x =  - {\sin ^2}\cos x\]

Observe that the substitution \(- \cos y = z\) will reduce this DE to a standard linear DE

\[\begin{align} - \cos y = z\\   \Rightarrow \qquad   \sin y\frac{{dy}}{{dx}} = \frac{{dz}}{{dx}} &  &  \ldots (2)\end{align}\]

Using (2) in (1), we have

\[\frac{{dz}}{{dx}} + \left( {2\cos x} \right)z =  - {\sin ^2}x\cos x\]

The I.F for this DE is \({e^{\int {2\cos xdx} }} = {e^{2\sin x}}\)

Thus, the solution will be given by

\[\begin{align}z{e^{2\sin x}} = - \int {{e^{2\sin x}}\cos x \cdot {{\sin }^2}x\,dx} \qquad\qquad ...(3)\end{align}\]

To integrate the R.H.S, we use the substitution \(\sin x = t\,\, \Rightarrow \,\,\cos xdx = dt.\) Thus, the integral reduces to

\[\begin{align}& - \int {{t^2}{e^{2t}}} \,dt\\& = - \frac{{{t^2}{e^{2t}}}}{2} - \frac{{{e^{2t}}}}{4} + \frac{{t{e^{2t}}}}{2} + C' &  \left( {{\rm{ Integration\;by\; parts}}} \right)\\& = - \frac{{{{\sin }^2}x \cdot {e^{2\sin x}}}}{2} - \frac{{{e^{2\sin x}}}}{4} + \frac{{\sin x \cdot {e^{2\sin x}}}}{2} + C'\end{align}\]

Finally, the solution to the DE is, from (3)

\[\begin{align}&z =  - \cos y =  - \frac{{{{\sin }^2}x}}{2} - \frac{1}{4} + \frac{{\sin x}}{2} + C'{e^{ - 2\sin x}}\\& \Rightarrow  \qquad 4\cos y = 2{\sin ^2}x - 2\sin x + 1 + C{e^{ - 2\sin x}}\end{align}\]

(b) Let \(\frac{{dy}}{{dx}} = p.\) Thus, this DE is

\[\begin{align}&\qquad\qquad{\left( {py + 2x} \right)^2} = \left( {1 + {p^2}} \right)\left( {{y^2} + 2{x^2}} \right)\\ &\Rightarrow  \quad 2{x^2}{p^2} - 4xyp + {y^2} - 2{x^2} = 0\\ &\Rightarrow  \quad p = \frac{{4xy \pm \sqrt {16{x^2}{y^2} - 8{x^2}\left( {{y^2} - 2{x^2}} \right)} }}{{4{x^2}}}\\& \qquad\quad= \frac{{4xy \pm \sqrt {8{x^2}{y^2} + 16{x^4}} }}{{4{x^2}}}\\& \qquad\quad = \frac{{4xy \pm 2\sqrt 2 \,x\sqrt {{y^2} + 2{x^2}} }}{{4{x^2}}}\\& \qquad\quad = \frac{y}{x} \pm \frac{{\sqrt {{y^2} + 2{x^2}} }}{{\sqrt 2 x}}\\ & \qquad\quad = \frac{y}{x} \pm \sqrt {\frac{1}{2}{{\left( {\frac{y}{x}} \right)}^2} + 1} \end{align}\]

The substitution y = vx reduces this DE to

\[\begin{align}&v + x\frac{{dv}}{{dx}} = v \pm \sqrt {\frac{{{v^2}}}{2} + 1} \\\Rightarrow \qquad&\frac{{dv}}{{\sqrt {{v^2} + 2} }} = \pm \frac{1}{{\sqrt 2 }}\frac{{dx}}{x}\\\Rightarrow \qquad & {\rm{Integrating}}\,{\rm{both}}\,{\rm{sides,}}\,{\rm{we}}\,{\rm{have}}\\&\ln \left| {v + \sqrt {{v^2} + 2} } \right| = \pm \frac{1}{{\sqrt 2 }}\ln x + C'\\ \Rightarrow \qquad &\ln x \pm \sqrt 2 \ln \left| {v + \sqrt {{v^2} + 2} } \right| = C\\ \Rightarrow \qquad &\ln x \pm \sqrt 2 \ln \left| {\frac{y}{x} + \frac{{\sqrt {{v^2} + 2{x^2}} }}{x}} \right| = C\qquad\qquad\ldots (4)\end{align}\]

Thus, we obtain two different solutions to the DE, one corresponding to the “ + ” and one to the “ – ” sign in (4).

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