Differential Equations Question 1
Question 1 - 2024 (01 Feb Shift 1)
Let $y=y(x)$ 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 :
(1) $\frac{4}{4+\sqrt{\mathrm{e}}}$
(2) $\frac{3}{3-\sqrt{\mathrm{e}}}$
(3) $\frac{2}{1+\sqrt{\mathrm{e}}}$
(4) $\frac{1}{2-\sqrt{\mathrm{e}}}$
Show Answer
Answer (4)
Solution
$\frac{d y}{d x}=2 x(x+y)^{3}-x(x+y)-1$
$x+y=t$
$\frac{d t}{d x}-1=2 x t^{3}-x t-1$
$\frac{d t}{2 t^{3}-t}=x d x$
$\frac{\mathrm{tdt}}{2 \mathrm{t}^{4}-\mathrm{t}^{2}}=\mathrm{xdx}$
Let $t^{2}=z$
$\int \frac{d z}{2\left(2 z^{2}-z\right)}=\int x d x$
$\int \frac{d z}{4 z\left(z-\frac{1}{2}\right)}=\int x d x$
$\ln \left|\frac{z-\frac{1}{2}}{z}\right|=x^{2}+k$
$z=\frac{1}{2-\sqrt{e}}$
