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

Marvin Mccormick

Marvin Mccormick

Answered question

2020-11-27

Solve differential equation y3y+1xy4=sinxx4

Answer & Explanation

Cullen

Cullen

Skilled2020-11-28Added 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

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Differential Equations

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?