Solve the initial value problem?

Question

y' + y = x, y(0) = 1.

then:
y'' − 7y' + 12y = 6x^2 + 5x − 18.

Answer 1

For the 1st equation, an easy solution is to multiply by an integration factor. Mutiplying both members by e^x, we get

e^x y' + e^x y = x e^x, and on the left hand side we have d/dx(y e^x). So, y e^x = Integral x e^x dx, which can be easily solved by parts, puting u = x and dv = e^x dx. Then

Integral x e^x dx = x e^x - Integral e^x dx = x e^x - e^x + C . It follows that y = x -1 + Ce^(-x).

Since y(0) = 0, 0 = -1 + C => C =1. So,

y = x -1 + e^(-x).

2) y'' − 7y' + 12y = 6x^2 + 5x − 18.

The solution of this equation isn split into 2 solutions, one, y_h, homogeneous, the other, y_p, particular. y_h is solution of

y'' − 7y' + 12y =0, and is given by the solutions of the algebraic equation r^2 - 7r + 12 = 0, which are r1 = 4 and r2 =3. Since they are real, then y_h = C1 e^(r1x) + C2 e^(r2x) = C1 e^(4x) + C2 e^(3x), where C1 and C2 are constants.

Since the right hand side is a polynomial f degree 2 , y_p is a polynomial of degree2, too. y_p = Ax^2 + Bx + C, where A, B and C are coefficientes to determine. Just plug y_p in the diffrerential equation. You get a linear system in A, B, C. Kinda boring but easy to solve

Answer 2

dy/dx + y = x. First solve
dy/dx + y = 0
dy/y = -dx
ln y + ln c = -x
ln cy = -x
cy = exp(-x)

y = C exp(-x)

For the particular solution, set y = A + Bx
dy/dx + y = B + (A + Bx) = x; evidently, B = 1, A = -B = -1

y = C exp(-x) - 1 + x

Checking,

dy/dx + y = [ -C exp(x) + 1 ] + [ C exp(-x) - 1 + x ] = x √


For the second problem...


y'' − 7y' + 12y = 6x^2 + 5x − 18. Solving first

d²y/dx² - 7 dy/dx + 12y = 0

Try y = C exp(mx), then the equation becomes

Cm² exp(mx) - 7Cm exp(mx) + 12C exp(mx) = 0

m² - 7m + 12 = 0
(m - 3)(m - 4) = 0; solutions are m = 3 and m = 4

So, y = C1 exp(3x) + C2 exp(4x)

For the particular solution, assume y = Ax² + Bx + D
Plugging this into the differential equation,

y'' − 7y' + 12y = 6x² + 5x − 18

2A - 7 (2Ax + B) + 12 (Ax² + Bx + D) = 6x² + 5x − 18

(12A - 6) x² + (-14A + 12B - 5) x + (2A -7B +12D + 18) = 0

12A - 6 = 0, so A = 1/2

-14A + 12B - 5 = -7 + 12B - 5 = 0, so B = 1

(2A - 7B + 12D + 18) = (1 - 7 + 12D + 18) = 0, so D = -1

The solution is

y = C1 exp(3x) + c2 exp(4x) + x² / 2 + x - 1

Answer 3

Homogeneous equation
y´+y=0 y= Ce^-x
Particular solution
y= ax+b
a+ax+b= x
a=1 b=-1 y =x-1
solution y=C e^-x+x-1 1= C-1 C=2
y=2e^-x +x-1