Let y=y(x) be the solution of the differential equation (dy)/(dx)=2x(x+y)^3-x(x+y)-1,y(0)=…

Question

Let $y=y(x)$ be the solution of the differential equation

$\frac{dy}{dx}=2x(x+y{)}^{3}-x(x+y)-1,y(0)=1$.

Then, ${(\frac{1}{\sqrt{2}}+y(\frac{1}{\sqrt{2}}))}^{2}$ equals :

Accepted Answer

$\begin{gathered} \frac{dy}{dx}=2x(x+y{)}^{3}-x(x+y)-1 \ x+y=t \ \Rightarrow \frac{dy}{dx}=\frac{dt}{dx}-1 \ \frac{dt}{dx}-1=2x(t{)}^{3}-xt \end{gathered}$

$\begin{gathered} \Rightarrow \frac{dt}{2{t}^{3}-t}=xdx \ \ \int \frac{1}{2{t}^{3}-t}dt=\int xdx \ \ \Rightarrow \int \frac{t}{2{t}^{4}-{t}^{2}}dt=\int xdx \end{gathered}$

$\begin{gathered} {t}^{2}=z \ 2tdt=dz \ \frac{1}{2}\int \frac{dz}{2{z}^{2}-z}=\int xdx \ \ln |\frac{z-\frac{1}{2}}{z}|={x}^{2}+c \end{gathered}$

$\ln |\frac{(x+y{)}^{2}-\frac{1}{2}}{(x+y{)}^{2}}|={x}^{2}+c$

$y(0)=1\Rightarrow c=\ln (\frac{1}{2})$

$\Rightarrow \frac{(x+y{)}^{2}-\frac{1}{2}}{(x+y{)}^{2}}={e}^{{x}^{2}}	imes \frac{1}{2}$

$\frac{(x+y{)}^{2}-\frac{1}{2}}{(x+y{)}^{2}}=\sqrt{e}	imes \frac{1}{2}$

$\Rightarrow (x+y{)}^{2}=\frac{1}{2-\sqrt{e}}$

Solutions · 1