Author Topic: TUT0601 Quiz1  (Read 920 times)

Joy Zhou

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TUT0601 Quiz1
« on: September 27, 2019, 11:35:07 PM »
Solve the given differential equation.

The given differential equation is

$
\frac{d y}{d x}=\frac{x-e^{-x}}{y+e^{y}}
$

i.e. $\left(y+e^{y}\right) \frac{d y}{d x}=\left(x-e^{-x}\right)$

or $\left(e^{-x}-x\right)+\left(y+e^{y}\right) \frac{d y}{d x}=0$

This equation is of the form $M(x)+N(y) \frac{d y}{d x}=0$ and hence is separable

On rewriting this equation we have

$\left(y+e^{y}\right) d y=\left(x-e^{-x}\right) d x$

On integrating $\int\left(y+e^{y}\right) d y=\int\left(x-e^{-x}\right) d x$

i.e. $\frac{y^{2}}{2}+e^{y}=\frac{x^{2}}{2}+e^{-x}+c^{\prime}$
 
where $c'$ is an arbitrary constant of integration

i.e. $y^{2}+2 e^{y}=x^{2}+2 e^{-x}+2 c^{\prime}$

Put $2 c^{\prime}=c$, another arbitrary

i.e. $y^{2}-x^{2}+2\left(e^{y}-e^{-x}\right)=c$ is the required solution

where $y+e^{y} \neq 0$