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# 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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