Verified Solution • 2025–26 Session moderate

QJEEM24Ma6647: JEE Main 2024 Mathematics - Class 12 JEEM24Ma6647 Solution: Let be the solution of the differential equation . Then, equals :

Let y=y(x)y=y(x) be the solution of the differential equation

dy dx=2x(x+y)3−x(x+y)−1,y(0)=1\frac{\mathrm{d} y}{\mathrm{~d} x}=2 x(x+y)^3-x(x+y)-1, y(0)=1 .

Then, (12+y(12))2\left(\frac{1}{\sqrt{2}}+y\left(\frac{1}{\sqrt{2}}\right)\right)^2 equals :

Multiple Choice Options

  • A 44+e\frac{4}{4+\sqrt{\mathrm{e}}}
  • B 33−e\frac{3}{3-\sqrt{\mathrm{e}}}
  • C 21+e\frac{2}{1+\sqrt{\mathrm{e}}}
  • D 12−e\frac{1}{2-\sqrt{\mathrm{e}}}

Step-by-Step Verified Solution

amp;dydx=2x(x+y)3−x(x+y)−1amp; Put x+y=tamp;⇒dydx=dtdx−1amp;dtdx−1=2x(t)3−xt\begin{aligned} & \frac{d y}{d x}=2 x(x+y)^3-x(x+y)-1 \\\\ & \text { Put } x+y=t \\\\ & \Rightarrow \frac{d y}{d x}=\frac{d t}{d x}-1 \\\\ & \frac{d t}{d x}-1=2 x(t)^3-x t\end{aligned}

⇒amp;dt2t3−t=xdxamp;∫12t3−tdt=∫xdx⇒amp;∫t2t4−t2dt=∫xdx\begin{aligned} \Rightarrow & \frac{d t}{2 t^3-t}=x d x \\\\ & \int \frac{1}{2 t^3-t} d t=\int x d x \\\\ \Rightarrow & \int \frac{t}{2 t^4-t^2} d t=\int x d x\end{aligned}

amp;t2=zamp;2tdt=dzamp;12∫dz2z2−z=∫xdxamp;ln⁡∣z−12z∣=x2+c\begin{aligned} & t^2=z \\\\ & 2 t d t=d z \\\\ & \frac{1}{2} \int \frac{d z}{2 z^2-z}=\int x d x \\\\ & \ln \left|\frac{z-\frac{1}{2}}{z}\right|=x^2+c\end{aligned}

ln⁡∣(x+y)2−12(x+y)2∣=x2+c\ln \left|\frac{(x+y)^2-\frac{1}{2}}{(x+y)^2}\right|=x^2+c

y(0)=1⇒c=ln⁡(12)y(0)=1 \Rightarrow c=\ln \left(\frac{1}{2}\right)

⇒(x+y)2−12(x+y)2=ex2×12\Rightarrow \frac{(x+y)^2-\frac{1}{2}}{(x+y)^2}=e^{x^2} \times \frac{1}{2}

(x+y)2−12(x+y)2=e×12\frac{(x+y)^2-\frac{1}{2}}{(x+y)^2}=\sqrt{e} \times \frac{1}{2}

⇒(x+y)2=12−e\Rightarrow(x+y)^2=\frac{1}{2-\sqrt{e}}

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