Need help with solving differentiation equation using integrating Factor

Question

The questions are:

dy/dx + 3y =e^-3x

dy/dx + ycotx = cosecx

x²dy/dx + xy = x +1

dy/dx - 3y/(x+1)=(x+1)⁴

Answer 1

1) y' + 3y = e^-3x

μ(x) = e^∫3dx = e^3x

e^3xy' + 3e^3xy = e^3x-3x = 1

(e^3xy)' = 1

e^3xy = x + C

y = (x+C)/e^3x

2) y' + ycot(x) = csc(x)

μ(x) = e^∫cot(x)dx = e^ln|sin(x)|=sin(x)

sin(x)y' + y cos(x) = 1

(sin(x)y)' = 1

sin(x)y = x+C

y = (x+C)/sin(x)

3) x²y' + xy = x+1

y' +y/x = 1/x + 1/x²

μ(x) = e^∫dx/x = e^ln|x| = x

xy'+ y = 1+1/x

(xy)' = 1+1/x

xy = x+ln|x|+ C

y = 1 + (ln|x|+C)/x

4) y'-3y/(x+1) = (x+1)⁴

μ(x) = e^{∫-3/(x+1)dx} = e^{ln|(x+1)^-3|} = 1/(x+1)³

y'/(x+1)³-3y/(x+1)⁴ = (y/(x+1)³)' = x+1

Integrating both sides:

y/(x+1)³ = x²/2+x+C

y = (x²/2+x+C)(x+1)³