उत्तर (4)

समाधान

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

$x+y=t$

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

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

$\Rightarrow\dfrac{tdt}{2 t^{4}-t^{2}}=xdx$

मान लीजिए $t^{2}=z$

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

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

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

$\Rightarrow z- \dfrac{1}{2} =zce^{x^{2}}$

$\Rightarrow z=\dfrac{1}{2(1+ce^{x^{2}})}$

$\Rightarrow t^{2}=\dfrac{1}{2(1+ce^{x^{2}})}$

$\Rightarrow (x+y)^{2}=\dfrac{1}{2(1+ce^{x^{2}})}$

जब $y=1$ तब $x=0$ है, तो हमें प्राप्त होता है,

$c=\dfrac{-1}{2}$

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

$\Rightarrow \left(\dfrac{1}{\sqrt{2}}+y\left(\dfrac{1}{\sqrt{2}}\right)\right)^{2}=\dfrac{1}{2-\sqrt{e}}$