Question

Solve the following IVP:

2. 2x^2 +y^2 +xyy′ =0, x>0, y(1)=1

Answer 1

Answer:

$x^2y^2+x^4=2$

Step-by-step explanation:

The given differential equation is:

$2x^2+y^2+xyy'=0,x\:>\:0,y(1)=1$

We rewrite this as

$2x^2+y^2+xy\frac{dy}{dx}=0,x\:>\:0,y(1)=1$

We make $\frac{dy}{dx}$ the subject to get:

$\frac{dy}{dx}=\frac{-2x^2-y^2}{xy}$....a homogeneous equation.

$\frac{dy}{dx}=-2(\frac{x}{y})-(\frac{y}{x})$

We use the following substitutions to obtain a seperable equation:

$y=vx,\frac{dy}{dx}=v+x\frac{dv}{dx},v=\frac{y}{x},\:and\:\frac{1}{v}=\frac{x}{y}$.

This implies that:

$v+\frac{xdv}{dx}=\frac{-2}{v}-v$

$\frac{xdv}{dx}=\frac{-2}{v}-2v$

$\frac{xdv}{dx}=\frac{-2-2v^2}{v}$

$\frac{v}{-2-2v^2}dv=\frac{dx}{x}$

We integrate both sides to obtain:

$\int \frac{v}{-2-2v^2}dv=\int \frac{dx}{x}$

$\frac{-\ln|2v^2+2|}{4}=\ln x+\ln K$

$-\ln|2v^2+2|=4(\ln x+\ln K)$

$\ln (\frac{1}{2v^2+2})=4\ln (Kx)$

$\ln (\frac{1}{2v^2+2})=\ln(K^4x^4)$

$\frac{1}{2v^2+2}=Cx^4$

$Cx^4(2v^2+2)=1$

We substitute $v^2=\frac{y^2}{x^2}$ to get:

$Cx^4(2*\frac{y^2}{x^2}+2)=1$

We substitute x=1,y=1 from the initial conditions.

$C(1)^4(2*\frac{1^2}{1^2}+2)=1$

$4C=1$

$C=\frac{1}{4}$

Our solution now becomes:

$x^2(2y^2+2x^2)=4$

$x^2(y^2+x^2)=2$

$x^2y^2+x^4=2$