Answer with explanation:
The given differential equation
y'y''=2--------(1)
We have to apply the following substitution
u=y'
u'=y"
Applying these substitution in equation (1)
u u'=2
[tex]u \frac{du}{dx}=2\\\\ u du=2 dx\\\\ \int u du=\int 2 dx\\\\\frac{u^2}{2}=2 x+K\\\\\frac{y'^2}{2}=2 x+K\\\\y'^2=4 x+2 K\\\\y'=(4 x+2 K)^{\frac{1}{2}}\\\\ dy=(4 x+2 K)^{\frac{1}{2}} d x\\\\\int dy=\int(4 x+2 K)^{\frac{1}{2}} d x\\\\y=\frac{(4 x+2 K)^{\frac{3}{2}}}{4 \times \frac{3}{2}}+J\\\\y=\frac{(4 x+2 K)^{\frac{3}{2}}}{6}+J[/tex]
Where , J and K are constant of Integration.
