QJEEM19Maeb0f: JEE Main 2019 Mathematics - Class 12 JEEM19Maeb0f Solution: The solution of the differential equation = x 2 (x 0) with y(1) …

Question

Verified Solution • 2025–26 Session moderate

The solution of the differential equation

x{{dy} \over {dx}} + 2y

= x² (x

\ne

0) with y(1) = 1, is :

Accepted Answer

x{{dy} \over {dx}} + 2y

= x²

\Rightarrow

{{dy} \over {dx}} + \left( {{2 \over x}} \right)y = x

\therefore

I.F =

{e^{\int {{2 \over x}dx} }}

= x²

\therefore

The solution is

yx² =

\int {{x^3}dx}

\Rightarrow

yx² =

{{{x^4}} \over 4} + C

.....(1)

As y(1) = 1

\therefore

when x = 1 then y = 1.

Putting the value of x and y in equation (1), we get

1 =

{1 \over 4} + C

\Rightarrow

C =

{3 \over 4}

\therefore

Required solution

yx² =

{{{x^4}} \over 4} + {3 \over 4}

\Rightarrow

y = {{{x^2}} \over 4} + {3 \over {4{x^2}}}

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