Verified Solution • 2025–26 Session moderate QJEEM24Mae915: JEE Main 2024 Mathematics - Class 12 JEEM24Mae915 Solution: Let be the solution of the differential equation . If , then is … Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (x2+4)2dy+(2x3y+8xy−2)dx=0(x^2+4)^2 d y+(2 x^3 y+8 x y-2) d x=0(x2+4)2dy+(2x3y+8xy−2)dx=0 . If y(0)=0y(0)=0y(0)=0 , then y(2)y(2)y(2) is equal to Multiple Choice Options A 2π2 \pi2π B π8\frac{\pi}{8}8π C π16\frac{\pi}{16}16π D π32\frac{\pi}{32}32π Step-by-Step Verified Solution dydx+y(2x3+8x)(x2+4)2=2(x2+4)2\frac{d y}{d x}+\frac{y\left(2 x^3+8 x\right)}{\left(x^2+4\right)^2}=\frac{2}{\left(x^2+4\right)^2}dxdy+(x2+4)2y(2x3+8x)=(x2+4)22 IF=e∫2x3+8x(x2+4)2dx\mathrm{IF}=e^{\int \frac{2 x^3+8 x}{\left(x^2+4\right)^2} d x}IF=e∫(x2+4)22x3+8xdx Let (x2+4)2=t⇒2(x2+4)(2x)dx=dt\text { Let }\left(x^2+4\right)^2=t \quad \Rightarrow 2\left(x^2+4\right)(2 x) d x=d t Let (x2+4)2=t⇒2(x2+4)(2x)dx=dt =e∫dt2t=elogt=t=(x2+4)∴y(x2+4)=∫2x2+4+c⇒y(x2+4)=tan−1(x2)+cy(0)=0\begin{gathered}=e^{\int \frac{d t}{2 t}}=e^{\log \sqrt{t}}=\sqrt{t}=\left(x^2+4\right) \\\therefore \quad y\left(x^2+4\right)=\int \frac{2}{x^2+4}+c \\\Rightarrow y\left(x^2+4\right)=\tan ^{-1}\left(\frac{x}{2}\right)+c \\y(0)=0\end{gathered}=e∫2tdt=elogt=t=(x2+4)∴y(x2+4)=∫x2+42+c⇒y(x2+4)=tan−1(2x)+cy(0)=0 amp;⇒0=0+c⇒c=0amp; put x=2amp;y(8)=π4⇒y=π32\begin{aligned}& \Rightarrow 0=0+c \quad \Rightarrow c=0 \\& \text { put } x=2 \\& y(8)=\frac{\pi}{4} \quad \Rightarrow \quad y=\frac{\pi}{32} \\\end{aligned}amp;⇒0=0+c⇒c=0amp; put x=2amp;y(8)=4π⇒y=32π Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring