How do i solve these hard maths questions?

Question

Solve the following initial value problem and give the domain of your solution. Check the initial value; you do not have to check the differential equation.

dy/dx+ y = 9xe^2x; with y = -1 when x = 0.

(b) Solve the following initial value problem and give the domain of your solution. Check the differential equation; you do not have to check the initial value.

dy/dx + cos(x)y = 2xe^((x^2)-sin(x)); with y = 0 when x = 0.

Answer 1

(a) Since the equation is already in standard form, the integrating factor is easily found:

e^[ ∫ dx] = e^x

Multiplying both sides by e^x results in

(d/dx)[ (e^x)y ] = 9xe^3x

Integrating both sides, where integration by parts is used on the right side, you have

(e^x)y  = c - e^(3x) + 3xe^(3x)

y = ce^x - e^(2x) + 3xe^(2x)

Using y(0) = -1,

-1 = c - 1 + 0

c = 0

So the solution is

y = - e^(2x) + 3xe^(2x)

The interval of solution is simply

-∞<x<∞

(b)

dy/dx + cos(x)y = 2xe^( x^2 - sin(x) )

Here the integrating factor is

e^[∫cos(x) dx ] = e^(sinx)

Multiplying both sides by  e^(sinx) results in

(d/dx)[ye^(sinx)] = 2xe^[ x^2 - sin(x) + sin(x) ]

(d/dx)[ ye^(sinx) ] = 2xe^[ x^2 ]

ye^(sinx) = e^(x^2) + c

y = e^[x^2 - sinx] + ce^(-sinx)

Using y(0)=0

0 = e^(0) + c

c = -1

So you have

y = e^[x^2 - sinx] - e^(-sinx)

The interval of solution is again

-∞<x<∞

Answer 2

dy/dx+ y = 9xe^2x

=> e^x dy/dx + y e^x = 9x e^3x

=> d/dx (ye^x) = 9xe^3x

=> ∫d(ye^x) = 9∫xe^3x dx

=> ye^x = 9[x∫e^3x dx - ∫[d/dx(x)∫e^3x dx]dx ]

=> ye^x = 3xe^3x - e^3x + c

Plugging x= 0 and y  -1

=> -1 = -1 + c => c = 0

=> y = (3x - 1)e^2x

Domain of y is R.

[I corrected the error in computing c after reading answer of Baker Street Irregular as above.]

(b)

dy/dx + cos(x)y = 2xe^((x^2)-sin(x))

=> e^sinx dy/dx + cosx * y * e^sinx = 2x e^(x²)

=> d/dx (ye^sinx) = 2xe^(x²)

=> d(ye^sinx) = 2xe^(x²) dx

=> ∫d(ye^sinx) =  2∫xe^(x²) dx

=> ye^sinx = e^(x²) + c

Plugging x=0 and y = 0

=> 0 = 1 + c => c = -1

=> y = [e^(x²) - 1] / e^sinx

Domain of y is R.