Let y=y(x) be the solution of the differential equation (x y-5 x^2 1+x^2 ) d x+(1+x^2) d y=0, y(0)=0 . Then y( 3 ) is equal to

Correct answer: B. 5 3 2 aligned & (1+x^2) (d y)/(d x)+x y=5 x^2 1+x^2 & (d y)/(d x)+(x y)/(1+x^2)= 5 x^2 1+x^2 & I.F. =e^ ∫ x 1+x^2 dx =e^ ( (1+x^2))/(2) = 1+x^2 & y 1+x^2 =∫ 5 x^2 1+x^2 · 1+x^2 dx & y 1+x^2 =∫ 5 x^2 1+x^2 · 1+x^2 dx & y 1+x^2 = 5 x^3 3 +C & y(0)=0 0=0+C C=0 aligned aligned & y= 5 x^3 3 1+x^2 & y( 3 )= 15 3 32 = 5 3 2 aligned

JEE Main 202524 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2025 · 24 January · Morning Shift · Mathematics · Differential Equations

Let y=y(x) be the solution of the differential equation (x y-5 x^2...

Let

\mathrm{y}=\mathrm{y}(\mathrm{x})

be the solution of the differential equation \left(x y-5 x^2 \sqrt{1+x^2}\right) d x+\left(1+x^2\right) d y=0, y(0)=0

. Then

y(\sqrt{3}) is equal to

Options

A $\sqrt{\frac{15}{2}}$
B $\frac{5 \sqrt{3}}{2}$
C $2 \sqrt{2}$
D $\sqrt{\frac{14}{3}}$

Correct answer

B. $\frac{5 \sqrt{3}}{2}$

Step-by-step solution

\begin{aligned} & \left(1+x^2\right) \frac{d y}{d x}+x y=5 x^2 \sqrt{1+x^2} \\ & \frac{d y}{d x}+\frac{x y}{1+x^2}=\frac{5 x^2}{\sqrt{1+x^2}} \\ & \therefore \text{ I.F. }=\mathrm{e}^{\int \frac{\mathrm{x}}{1+x^2} \mathrm{dx}}=\mathrm{e}^{\frac{\ln \left(1+x^2\right)}{2}}=\sqrt{1+\mathrm{x}^2} \\ & \therefore \mathrm{y} \sqrt{1+\mathrm{x}^2}=\int \frac{5 \mathrm{x}^2}{\sqrt{1+\mathrm{x}^2}} \cdot \sqrt{1+\mathrm{x}^2} \mathrm{dx} \\ & \therefore \mathrm{y} \sqrt{1+\mathrm{x}^2}=\int \frac{5 \mathrm{x}^2}{\sqrt{1+\mathrm{x}^2}} \cdot \sqrt{1+\mathrm{x}^2} \mathrm{dx} \\ & \mathrm{y} \sqrt{1+\mathrm{x}^2}=\frac{5 \mathrm{x}^3}{3}+\mathrm{C} \\ & \because \mathrm{y}(0)=0 \Rightarrow 0=0+\mathrm{C} \Rightarrow \mathrm{C}=0\end{aligned}

\begin{aligned} & \therefore y=\frac{5 x^3}{3 \sqrt{1+x^2}} \\ & y(\sqrt{3})=\frac{15 \sqrt{3}}{32}=\frac{5 \sqrt{3}}{2}\end{aligned}

Practice Differential Equations on Quantrex Academy →

More from Differential Equations