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

Marvin Mccormick 2020-11-27 Answered

Solve differential equation y3y+1xy4=sinxx4

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Expert Answer

Cullen
Answered 2020-11-28 Author has 89 answers

Let y4=z then, 4y3y=dzdx
y3y+1xy4=sinxx414dzdx+1xz=sinxx4
dzdx+4xz=4sinxx4
The given ordinary equation is of the form
dzdx+P(x)z=Q(x)
Integrating factor: IF=ePdx=e4xdx=x4
ddx(IFz)=IFQ(x)
IFz=IF1(x)dx
x4z=(x4)(4sinxx)dx
Multiply the original differential equation by the integral factor
ddx(IFz)=IFQ(x)
IFz=IFQ(x)dx
x4z=(x4)(4sinxx4)dx
x4z=(4sinx)dx
x4z=4sinxdx
x4z=4cosx+c
z=4cosxx4+cx4
Put the value of z in the obtained equation
z=4cosxx4+cx4
z=y4
y4=4cosxx4+cx4
y=(4cosxx4+cx4)14

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