Question

The solution of the differential equation $y^{3}dy+(x+y^2)dx=0$ is:

• A. $\frac{1}{2}log(x^4+2x^{2}y+2y^2)+K$
• B. $\frac{1}{2}log(y^4+2y^{2}x+2x^2)+K$
• C. $log(y^4+2x^2+2x{y}^2)-2ta{n}^{-1}\left(\frac{y^2+x}{x}\right)+K$
• D. None of these

Answer 1

Answer: (C)

$log(y^4+2x^2+2x{y}^2)-2ta{n}^{-1}\left(\frac{y^2+x}{x}\right)+K$

$y^{3}dy+(x+y^2)dx=0$

Let $y^2=z$

$2y.dy=dz$

$\frac{dz}{2\sqrt{z}}=dy$

Substituting we get

$z\sqrt{z}\frac{dz}{2\sqrt{z}}+(x+z)dx=0$

$zdz=-2(x+z)dx$

Let $z=vx$

$dz=xdv+vdx$

Substituting we get

$vx(xdv+vdx)=-2(x+vx)dx$

$v(xdv+vdx)=-2(1+v)dx$

$\frac{vxdv}{dx}+v^2=-2-2v$

$v^2+2v+2=-\frac{vxdv}{dx}$

$(v+1{)}^2+1=-\frac{vxdv}{dx}$

${\int }_{\frac{dx}{x}}=-{\int }_{(}\frac{vdv}{(v+1{)}^2+1)})$

$lnx+c=-{\int }_{(}\frac{(v+1-1)dv}{(v+1{)}^2+1)})$

$lnx+c=-{\int }_{\frac{v+1\left(dv\right)}{(v+1{)}^2+1)}}+{\int }_{\frac{dv}{(v+1{)}^2+1)}}$

$lnx+c=-\frac{ln\left(\right(v+1{)}^2+1)}{2}+ta{n}^{-1}(v+1)$

$2lnx+C=2ta{n}^{-1}(v+1)-\frac{ln\left(\right(v+1{)}^2+1)}{2}$

$2lnx+C=2ta{n}^{-1}\left(\frac{x+y^2}{x}\right)-ln\left(\right(x+y^2{)}^2+x^2)+2lnx$

$C=2ta{n}^{-1}\left(\frac{x+y^2}{x}\right)-ln(y^4+2x^2+2x{y}^2)$