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