Answer to Question #216743 in Differential Equations for Unknown346307

Question #216743

xydy/dx+4x2+y2=0 where y(2)=-7,x>0


1
Expert's answer
2021-07-19T10:50:17-0400

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 eln=1;

(2v2+4)×(x4)=C3;where C3="e^{C_2}"

But "v=\\frac yx"

"(2\\frac{y^2}{x^2}+4)\u00d7(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








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