Integration using integrating factor

[xlalcciax]

Homework Statement

Integrate dy/dx=2y+4x+10

The Attempt at a Solution

dy/dx-2y=4x+10
Integrating factor = e^(-2)dx=e^-2x
multiply both sides by IF. (e^-2x)dy/dx-2y(e^-2x)=(e^-2x)(4x+10)
dy/dx(e^-2x y)=(e^-2x)(4x+10)
i don't know what to do next.

 

[Mark44]

Homework Statement

Integrate dy/dx=2y+4x+10

The Attempt at a Solution

dy/dx-2y=4x+10
Integrating factor = e^(-2)dx=e^-2x
multiply both sides by IF. (e^-2x)dy/dx-2y(e^-2x)=(e^-2x)(4x+10)
dy/dx(e^-2x y)=(e^-2x)(4x+10)
i don't know what to do next.

The left side is actually d/dx(ye^-2x), which is different from dy/dx(ye^-2x).

The equation is d/dx(ye^-2x) = (4x + 10)e^-2x.
Integrate both sides with respect to x, which gives you
ye^{-2x} = \int (4x + 10)e^{-2x}dx

Can you take it from here?

 

[xlalcciax]

The left side is actually d/dx(ye^-2x), which is different from dy/dx(ye^-2x).

The equation is d/dx(ye^-2x) = (4x + 10)e^-2x.
Integrate both sides with respect to x, which gives you
ye^{-2x} = \int (4x + 10)e^{-2x}dx

Can you take it from here?

i know that the left side is ye^-2x but i don't know how to integrate (4x + 10)e^{-2x}

 

[Mark44]

Split the right-side integral into two integrals. For one of them, use integration by parts. The other one is pretty easy.