Let y = y ( x ) be the solution of the differential equation d y/d x=2 x(x+y) 3 -x(x+y)-1, y(0)=1 . Then, (frac 1 √(2) + y (frac 1 √(2) ) ) 2 equals :

Mathematics · Differential Equations

JEE Main 2024 — 1 February, Shift 1 — Question 14

Let $\mathrm{y}=\mathrm{y}(\mathrm{x})$ be the solution of the differential equation

$\frac{d y}{d x}=2 x(x+y)^{3}-x(x+y)-1, y(0)=1$ . Then, $\left(\frac{1}{\sqrt{2}}+\mathrm{y}\left(\frac{1}{\sqrt{2}}\right)\right)^{2}$ equals :

$\frac{d y}{d x}=2 x(x+y)^{3}-x(x+y)-1$

$x+y=t$

$\frac{d t}{d x}-1=2 x t^{3}-x t-1$

$\frac{d t}{2 t^{3}-t}=x d x$

$\frac{t d t}{2 t^{4}-t^{2}}=x d x$

Let $t^{2}=z$

$\int \frac{d z}{2\left(2 z^{2}-z\right)}=\int x d x$

$\int \frac{d z}{4 z\left(z-\frac{1}{2}\right)}=\int x d x$

$\ln \left|\frac{z-\frac{1}{2}}{z}\right|=x^{2}+k$

$z=\frac{1}{2-\sqrt{e}}$

Answer key and solution verified before publishing.

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