Question

xydy/dx+4x²+y²=0 where y(2)=-7,x>0

Answer 1

Solution

xy$\frac{dy}{dx}+4x^2+y^2=0$

Rewrite as follows;

$\frac{dy}{dx}+\frac{4x^2+y^2}{xy}=0$

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

Substitute in the equation;

$v+x\frac{dv}{dx}+\frac{4x^2+v^2x^2}{vx^2}=0$

Simplify;

$v+x\frac{dv}{dx}+\frac{4+v^2}{v}=0$

Compute by variables;

$x\frac{dv}{dx}+\frac{4+2v^2}{v}=0$

Seperate by variables;

$\frac{v}{4+2v^2}dv+\frac{1}{x}dx=0$

Integrate;

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

Which can be simplified as;

ln(2v²+4)+4ln(x)=C₂ ,where C₂=4C

Can be further simplified and written as;

ln[(2v²+4)×(x⁴)]=C₂

Remove ln with e,since e^ln=1;

(2v²+4)×(x⁴)=C₃;where C₃=$e^{C_2}$

But $v=\frac yx$

$(2\frac{y^2}{x^2}+4)×(x^4)=C_3$

Simplify;

y²=$(\frac{C_3-4x^4}{2x^2})$ ,x>0

Apply the initial condition;

y(2)=-7

(-7)²=$\frac{C_3-64}{8}$

C₃=456

y=$-\sqrt{\frac{456-4x^2}{2x^2}}$ ;x>0