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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
For differential equation
$y e^{\frac{x}{y}},dx=(x e^{\frac{x}{y}}+y^2),dy$,
$y(0)=1$, the value of $x(e)$ is equal to:
Solution
Given
$y e^{\frac{x}{y}},dx=(x e^{\frac{x}{y}}+y^2),dy$
Divide by $dy$:
$y e^{\frac{x}{y}}\frac{dx}{dy}=x e^{\frac{x}{y}}+y^2$
Notice that
$\frac{d}{dy}\left(x e^{\frac{x}{y}}\right)=y e^{\frac{x}{y}}\frac{dx}{dy}-x e^{\frac{x}{y}}$
So equation becomes
$\frac{d}{dy}\left(x e^{\frac{x}{y}}\right)=y^2$
Integrate:
$x e^{\frac{x}{y}}=\frac{y^3}{3}+C$
Now use initial condition $y(0)=1$
So at $x=0$, $y=1$
$0\cdot e^{0}= \frac{1}{3}+C$
$C=-\frac{1}{3}$
Thus,
$x e^{\frac{x}{y}}=\frac{y^3-1}{3}$
Now put $y=e$
$x e^{\frac{x}{e}}=\frac{e^3-1}{3}$
It satisfies $x=1$
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