69. <math><mrow><mrow><mo>(</mo><mrow><mn>1</mn><mo>+</mo><msup><mrow><mi>e</mi></mrow><mrow><mfrac><mrow><mi>x</mi></mrow><mrow><mi>y</mi></mrow></mfrac></mrow></msup></mrow><mo>)</mo></mrow><mi>d</mi><mi>x</mi><mo>+</mo><msup><mrow><mi>e</mi></mrow><mrow><mfrac><mrow><mi>x</mi></mrow><mrow><mi>y</mi></mrow></mfrac></mrow></msup><mrow><mo>(</mo><mrow><mn>1</mn><mo>−</mo><mfrac><mrow><mi>x</mi></mrow><mrow><mi>y</mi></mrow></mfrac></mrow><mo>)</mo></mrow><mi>d</mi><mi>y</mi><mo>=</mo><mn>0</mn></mrow></math>

The given D.E. is(1+exy)dx+exy(1−xy)dy=0⇒(1+exy)dx=−exy(1−xy)dy⇒dxdy=−exy(1−xy)1+exy=f(xy)Hence, the given D.E. is homogenous.Let, x=yv=yx so that dydx=v+ydxdy in the D.E.Then, v+ydvdy=−ev(1−v)1+ev⇒ydvdy=vev−ev1+ev−v=vev−ev−v−vev1+ev⇒ydvdy=−(ev+v)1+ev⇒(1+evev+v)dv=−dyyIntegrating both sides we get,log|ev+v|=−log|y|+log|c|Putting back v=xy we get,log|exy+xy|=log|cy|=exy+xy=cy=x+yexy=c is the general solution.

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Differential Equations Ncert Solutions Maths class 12th Maths Class 12th Differential Equations

69. $(1 + e^{\frac{x}{y}}) d x + e^{\frac{x}{y}} (1 - \frac{x}{y}) d y = 0$

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Vishal Baghel | Contributor-Level 10

4 months ago

The given D.E. is

$\begin{array}{l} (1 + e^{\frac{x}{y}}) d x + e^{\frac{x}{y}} (1 - \frac{x}{y}) d y = 0 \\ \Rightarrow (1 + e^{\frac{x}{y}}) d x = - e^{\frac{x}{y}} (1 - \frac{x}{y}) d y \\ \Rightarrow \frac{d x}{d y} = - \frac{e^{\frac{x}{y}} (1 - \frac{x}{y})}{1 + e^{\frac{x}{y}}} = f (\frac{x}{y}) \end{array}$

Hence, the given D.E. is homogenous.

Let, $x = y v = \frac{y}{x}$ so that $\frac{d y}{d x} = v + y \frac{d x}{d y}$ in the D.E.

Then, $v + y \frac{d v}{d y} = \frac{- e^{v} (1 - v)}{1 + e^{v}}$

$\begin{array}{l} \Rightarrow y \frac{d v}{d y} = \frac{v e^{v} - e^{v}}{1 + e^{v}} - v = \frac{v e^{v} - e^{v} - v - v e^{v}}{1 + e^{v}} \\ \Rightarrow y \frac{d v}{d y} = - \frac{(e^{v} + v)}{1 + e^{v}} \\ \Rightarrow (\frac{1 + e^{v}}{e^{v} + v}) d v = - \frac{d y}{y} \end{array}$

Integrating both sides we get,

$l o g | e^{v} + v | = - l o g | y | + l o g | c |$

Putting back $v = \frac{x}{y}$ we get,

$\begin{array}{l} l o g | e^{\frac{x}{y}} + \frac{x}{y} | = l o g | \frac{c}{y} | \\ = e^{\frac{x}{y}} + \frac{x}{y} = \frac{c}{y} \end{array}$

$= x + y e^{\frac{x}{y}} = c$ is the general solution.

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