The Solution of differential equation dx = e x - y + x 2 e - y is

Given that, dx = e x - y + x 2 e - y dx = e x e - y + x 2 e - y dx = + x 2 e y dy = ( e x + x 2 )dx On integrating both sides, we get dy = + x 2 ) dx = e x + 3 + C - e x = 3 + C

Differential Equations — Class 12 Maths Solution

exemplar objective MCQ NCERT EXEMP.Q.74,Page.201

Question

The Solution of differential equation $\frac{{dy}}{{dx}} = {e^{x - y}} + {x^2}{e^{ - y}}$ is

(a) $y = {e^{x - y}} - {x^2}{e^{ - y}} + C$
(b) ${e^y} - {e^x} = \frac{{{x^3}}}{3} + C$ ✓ Correct
(c) ${e^x} + {e^y} = \frac{{{x^3}}}{3} + C$
(d) ${e^x} - {e^y} = \frac{{{x^3}}}{3} + C$

Step-by-step Solution

Given that,

$\frac{{dy}}{{dx}} = {e^{x - y}} + {x^2}{e^{ - y}}$

$\Rightarrow$ $\frac{{dy}}{{dx}} = {e^x}{e^{ - y}} + {x^2}{e^{ - y}}$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{{e^x} + {x^2}}}{{{e^y}}}$

$\Rightarrow$ ${e^y}dy = \left( {{e^x} + {x^2}} \right)dx$

On integrating both sides,

we get
$\int {{e^y}} dy = \int {\left( {{e^x} + {x^2}} \right)} dx$

$\Rightarrow$ ${e^y} = {e^x} + \frac{{{x^3}}}{3} + C$

$\Rightarrow$ ${e^y} - {e^x} = \frac{{{x^3}}}{3} + C$

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