Solve the following IVP:2. 2x^2 +y^2 +xyy′ =0, x>0, y(1)=1

Answer:[tex]x^2y^2+x^4=2[/tex]Step-by-step explanation:The given differential equation is:[tex]2x^2+y^2+xyy'=0,x\:>\:0,y(1)=1[/tex]We rewrite this as[tex]2x^2+y^2+xy\frac{dy}{dx}=0,x\:>\:0,y(1)=1[/tex]We make [tex]\frac{dy}{dx}[/tex] the subject to get:[tex]\frac{dy}{dx}=\frac{-2x^2-y^2}{xy}[/tex]....a homogeneous equation.[tex]\frac{dy}{dx}=-2(\frac{x}{y})-(\frac{y}{x})[/tex]We use the following substitutions to obtain a seperable equation:[tex]y=vx,\frac{dy}{dx}=v+x\frac{dv}{dx},v=\frac{y}{x},\:and\:\frac{1}{v}=\frac{x}{y}[/tex].This implies that:[tex]v+\frac{xdv}{dx}=\frac{-2}{v}-v[/tex][tex]\frac{xdv}{dx}=\frac{-2}{v}-2v[/tex][tex]\frac{xdv}{dx}=\frac{-2-2v^2}{v}[/tex][tex]\frac{v}{-2-2v^2}dv=\frac{dx}{x}[/tex]We integrate both sides to obtain:[tex]\int \frac{v}{-2-2v^2}dv=\int \frac{dx}{x}[/tex][tex]\frac{-\ln|2v^2+2|}{4}=\ln x+\ln K[/tex][tex]-\ln|2v^2+2|=4(\ln x+\ln K)[/tex][tex]\ln (\frac{1}{2v^2+2})=4\ln (Kx)[/tex][tex]\ln (\frac{1}{2v^2+2})=\ln(K^4x^4)[/tex][tex]\frac{1}{2v^2+2}=Cx^4[/tex][tex]Cx^4(2v^2+2)=1[/tex]We substitute [tex]v^2=\frac{y^2}{x^2}[/tex] to get:[tex]Cx^4(2*\frac{y^2}{x^2}+2)=1[/tex]We substitute x=1,y=1 from the initial conditions.[tex]C(1)^4(2*\frac{1^2}{1^2}+2)=1[/tex][tex]4C=1[/tex][tex]C=\frac{1}{4}[/tex]Our solution now becomes:[tex]x^2(2y^2+2x^2)=4[/tex][tex]x^2(y^2+x^2)=2[/tex][tex]x^2y^2+x^4=2[/tex]