JEE MAIN - Mathematics (2024 - 1st February Morning Shift - No. 14)
Let $y=y(x)$ be the solution of the differential equation
$\frac{\mathrm{d} y}{\mathrm{~d} x}=2 x(x+y)^3-x(x+y)-1, y(0)=1$.
Then, $\left(\frac{1}{\sqrt{2}}+y\left(\frac{1}{\sqrt{2}}\right)\right)^2$ equals :
$\frac{\mathrm{d} y}{\mathrm{~d} x}=2 x(x+y)^3-x(x+y)-1, y(0)=1$.
Then, $\left(\frac{1}{\sqrt{2}}+y\left(\frac{1}{\sqrt{2}}\right)\right)^2$ equals :
$\frac{4}{4+\sqrt{\mathrm{e}}}$
$\frac{3}{3-\sqrt{\mathrm{e}}}$
$\frac{2}{1+\sqrt{\mathrm{e}}}$
$\frac{1}{2-\sqrt{\mathrm{e}}}$
Explanation
$\begin{aligned} & \frac{d y}{d x}=2 x(x+y)^3-x(x+y)-1 \\\\ & \text { Put } x+y=t \\\\ & \Rightarrow \frac{d y}{d x}=\frac{d t}{d x}-1 \\\\ & \frac{d t}{d x}-1=2 x(t)^3-x t\end{aligned}$
$\begin{aligned} \Rightarrow & \frac{d t}{2 t^3-t}=x d x \\\\ & \int \frac{1}{2 t^3-t} d t=\int x d x \\\\ \Rightarrow & \int \frac{t}{2 t^4-t^2} d t=\int x d x\end{aligned}$
$\begin{aligned} & t^2=z \\\\ & 2 t d t=d z \\\\ & \frac{1}{2} \int \frac{d z}{2 z^2-z}=\int x d x \\\\ & \ln \left|\frac{z-\frac{1}{2}}{z}\right|=x^2+c\end{aligned}$
$\ln \left|\frac{(x+y)^2-\frac{1}{2}}{(x+y)^2}\right|=x^2+c$
$y(0)=1 \Rightarrow c=\ln \left(\frac{1}{2}\right)$
$\Rightarrow \frac{(x+y)^2-\frac{1}{2}}{(x+y)^2}=e^{x^2} \times \frac{1}{2}$
$\frac{(x+y)^2-\frac{1}{2}}{(x+y)^2}=\sqrt{e} \times \frac{1}{2}$
$\Rightarrow(x+y)^2=\frac{1}{2-\sqrt{e}}$
$\begin{aligned} \Rightarrow & \frac{d t}{2 t^3-t}=x d x \\\\ & \int \frac{1}{2 t^3-t} d t=\int x d x \\\\ \Rightarrow & \int \frac{t}{2 t^4-t^2} d t=\int x d x\end{aligned}$
$\begin{aligned} & t^2=z \\\\ & 2 t d t=d z \\\\ & \frac{1}{2} \int \frac{d z}{2 z^2-z}=\int x d x \\\\ & \ln \left|\frac{z-\frac{1}{2}}{z}\right|=x^2+c\end{aligned}$
$\ln \left|\frac{(x+y)^2-\frac{1}{2}}{(x+y)^2}\right|=x^2+c$
$y(0)=1 \Rightarrow c=\ln \left(\frac{1}{2}\right)$
$\Rightarrow \frac{(x+y)^2-\frac{1}{2}}{(x+y)^2}=e^{x^2} \times \frac{1}{2}$
$\frac{(x+y)^2-\frac{1}{2}}{(x+y)^2}=\sqrt{e} \times \frac{1}{2}$
$\Rightarrow(x+y)^2=\frac{1}{2-\sqrt{e}}$
Comments (0)
