Let $y=y(x)$ be the solution of the differential equation $\dfrac{dy}{dx}=2x(x+y)^3-x(x+y)-1$, $y(0)=1$. Then $\left(\dfrac{1}{\sqrt{2}}+y\left(\dfrac{1}{\sqrt{2}}\right)\right)^2$ equals:
Step-by-Step Solution
Key Concept: Substitute $t=x+y$ so $\frac{dt}{dx}-1=2xt^3-xt-1$, giving $\frac{dt}{dx}=xt(2t^2-1)$. Separate variables, let $z=t^2$, integrate to get $\ln\left|\frac{z-1/2}{z}\right|=x^2+k$. Apply IC to find $z$.
Let $t=x+y$: $\frac{dt}{dx}=xt(2t^2-1)$. Let $z=t^2$: $\int\frac{dz}{2z(2z-1)}=\int x\,dx\Rightarrow\ln\left|\frac{z-1/2}{z}\right|=x^2+k$. IC $y(0)=1\Rightarrow t(0)=1,z(0)=1$: $k=\ln(1/2)$. At $x=\frac{1}{\sqrt{2}}$: $t^2=(x+y)^2=\frac{1}{2-\sqrt{e}}$.
Correct Answer: 4