Solve differential equation y^3y'+1/x y^4= sinx/x^4

Solve differential equation y^3y'+1/x y^4= sinx/x^4

Question
Solve differential equation \(y^3y'+1/x y^4= sinx/x^4\)

Answers (1)

2020-11-28
Let \(y^4= z\) then, \(4y^3 y'= dz/dx\)
\(y^3y'+1/x y^4= sinx/x^4 => 1/4 dz/dx+1/x z=sinx/x^4\)
\(dz/dx+4/x z= (4sinx)/x^4\)
The given ordinary equation is of the form
\(dz/dx+P(x)z=Q(x)\)
Integrating factor: \(IF= e^(int Pdx)= e(int 4/x)dx= x^4\)
\(d/dx(IFz)=IFQ(x)\)
\(IF*z= int IF*1(x)dx\)
\(x^4*z= int(x^4)((4sinx)/x)dx\)
Multiply the original differential equation by the integral factor
\(d/dx(IF*z)= IF*Q(x)\)
\(IF*z= int IF*Q(x)dx\)
\(x^4*z=int (x^4)((4sinx)/x^4)dx\)
\(x^4*z= int (4sinx)dx\)
\(x^4z= 4 int sin xdx\)
\(x^4z= -4cosx+c\)
\(z= (-4cosx)/x^4+c/x^4\)
Put the value of z in the obtained equation
\(z= (-4cosx)/x^4+c/x^4\)
\(z= y^4\)
\(y^4= (-4cosx)/x^4+c/x^4\)
\(y= root(1/4)(((-4cosx)/x^4+c/x^4))\)
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