Question

/10 POINT ZILLDIFFEQMODAP11 4.4.003. Solve the given differential equation by undetermined coefficients. y" – 4y' + 4y = 4x + 4 y(x) = Submit Answer

Answer 1

Undetermined coefficeients method

We can solve the given initial value problem by using method of variation of parameters method.

So the most general solution is Y = YC + YP

YC = Homogeneous solutin

YP = Particular solution.

Now the given initil value problem is Y’’- 4Y’ + 4Y = 4X + 4

So let us take the homogeneous equation for solving

So Y’’- 4Y’ + 4Y = 0 -------------------(1)

Now let us find the characteristic equation for homogeneous equation by assuming the solution y = erx not equla to Zero.

So y’ = r erx

      y’’ = r2erx

so substitute in (1)

r2erx - 4rerx + 4erx = 0

erx (r2 - 4r + 4 ) = 0

erx is not equal to zero so r2 - 4r + 4 = 0

So r2 - 2r – 2r + 4 = 0

So r(r – 2) -2(r – 2) = 0

So (r - 2)(r - 2) = 0

So r1 = r2 = 2

So the roots of the characterestic equation is real and repeat values then the general solution is in the form of Yc = C1er1t + C2ter2t

So Yc = C1e2t + C2te2t

Now for a particular solution

We have a non homogeneous term g(x) = 4x + 4

Now the differential cycle g’(x) is = 4

And g’’(x) = 0

so we can assume Yp = AX + B

then Y’p = A

Y’’p = 0

Now substitute in given IVP

• Yp’’- 4Yp’ + 4Yp = 4X + 4
• 0 – 4A + 4(AX + B) = 4X + 4
• -4A + 4B + 4AX = 4X + 4

Now equal x coefficients both sides

So 4A = 4 then A = 1

Now equal constant terms

Then -4A + 4B = 4

Then –A + B = 1

Substitute A value the -1 + B = 1 then B = 1+1 = 2

So B = 2

So particular solution = Yp = AX + B = X + 2

So the most general solution Y = Yc + Yp

So Y = C1e2t + C2te2t + x + 2