Question

Find the using variation general solution of parameters y"^y'-2y=2et Y

Answer 1

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

So the most general solution is   Y = Y_C + Y_P

Y_C = Homogeneous solutin

Y_P = Particular solution.

Now the given initil value problem is y’’ - y’ - 2y = 2e^-t

So let us take the homogeneous equation for solving

So y’’ - y’ - 2y = 0    -------------------(4)

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

So y’ = r e^rt

And y’’ = r²e^rt

so substitute in (1)

r²e^rt - re^rt - 2e^rt = 0

e^rt (r² - r - 2) = 0

e^rt is not equal to zero so r² - r - 2 = 0

so r² - r - 2 = 0

r² + r - 2r - 2 = 0

r (r+ 1) - 2(r + 1) = 0

(r - 2) (r + 1) = 0

r₁ = 2 and r₂ = -1

so when the characteristic equation roots are real and differrent then homogeneous solution is in the form as given below.

Yc = C₁e^r1t + C₂e^r2t

Yc = C₁e^2t + C₂e^-t

So now we need to find the wronskian for particular solution

Yc = C₁y₁(t) + C₂y₂(t)

So y₁(t) = e^2t and y₂(t) = e^-t

y₁’(t) = 2e^2t and y₂’(t) = -e^-t

And  W(y₁ , y₂) = y₁(t) y₂’(t) - y₁’(t) y₂(t)

W(y₁ , y₂) = (e^2t) (-e^-t) – (2e^2t) (e^-t)

W(y₁ , y₂) = -(e^2t-t) – (2e^2t-t)

W(y₁ , y₂) = -(e^t) – (2e^t)

W(y₁ , y₂) =  – (3e^t)

And non homogeneous term g(t) = 2e^-t

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