QJEEM19Ma5beb: JEE Main 2019 Mathematics - Class 12 JEEM19Ma5beb Solution: If y = y(x) is the solution of the differential equation, x +…

Question

Verified Solution • 2025–26 Session moderate

If y = y(x) is the solution of the differential equation,

x

dy \over dx

+ 2y = x², satisfying y(1) = 1, then y(

1\over2

) is equal to :

Accepted Answer

Given,

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

\Rightarrow

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

This is a linear differential equation.

\therefore

I.F

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

= {e^{2\ln x}}

= {x^2}

\therefore

Solution is,

y \cdot {x^2} = \int {x \cdot {x^2}dx}

\Rightarrow

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

given

y\left( 1 \right) = 1

\therefore

1.1 = {4 \over 4} + C

\Rightarrow

C = {3 \over 4}

\therefore

Equation is

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

\therefore

y\left( {{1 \over 2}} \right)

means

x = {1 \over 2}

\therefore

y \cdot {\left( {{1 \over 2}} \right)^2} = {1 \over 4} \times {\left( {{1 \over 2}} \right)^4} + {3 \over 4}

\Rightarrow

{y \over 4} = {1 \over {64}} + {3 \over 4}

\Rightarrow

{y \over 4} = {{1 + 48} \over {64}}

\Rightarrow

y =

{{49} \over {16}}

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