JEE Advanced 2026 · Paper 2, Mathematics question 3

Separate.
$$\frac{1+y^4}{y^3}\,dy=\frac{e^{5x}+1}{e^x}\,dx=\left(e^{4x}+e^{-x}\right)dx.$$
Integrating,
$$\frac{y^2}2-\frac1{2y^2}=\frac{e^{4x}}4-e^{-x}+C.$$
At $x=0$, $y^2=\frac12$: $\frac14-1=\frac14-1+C$, so $C=0$.

At $x=\ln2$. The right side is $4-\frac12=\frac72$, so $y^4-7y^2-1=0$ and $y^2=\dfrac{7+\sqrt{53}}2$ (the positive root). Since $y>0$,
$$y(\ln2)=\sqrt{\frac{7+\sqrt{53}}2}.$$