The solution $y^{n-1}\left(a y^{\prime}+y\right)=x$ is

$(n-1) y^{n-1}=C e^{-\frac{n x}{a}}+n x+a$

$n y^{n-1}=C e^{-\frac{n x}{a}}+n x$

$n y^{n-1}=C e^{-\frac{n x}{a}}+n x+a$

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

The solution

y^{n-1}\left(a y^{\prime}+y\right)=x

View answer and explanation

Correct option: C

n y^n=C e^{-\frac{n x}{a}}+n x-a

Option C (

n y^n=C e^{-\frac{{n x}}{{a}}}+n x-a

) is correct because it is the solution obtained by making the substitution

u = y^n

and solving the resulting linear differential equation.

Practice All Chapter MCQs →

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\lim \limits_{{x \rightarrow \infty}} y(t)