Answer to Question #280247 in Differential Equations for Catherine

Question

y''-4y'+4y=(x+1)e^x

Answer 1

y" - 4y'+4y = (x+1)e^x

For complementary function , the auxiliary equation is

m² - 4m + 4 = 0

=> m² - 2.m.2 + 2² = 0

=> (m-2)² = 0

=> m = 2, 2

So the complementary function is given by

C.F. = (c₁+c₂x)e^2x

Now let's find particular integral

P.I. = $\frac{1}{f(D)}(x+1)e^{x}$ , where f(D) = (D-2)², Dⁿ = $\frac{d^{n}}{dx^{n}}$

So P.I. = e^x$\frac{1}{f(D+1)}(x+1)$

= e^x $\frac{1}{(D+1-2)²}(x+1)$

= e^x$\frac{1}{(D-1)²}(x+1)$

= e^x$\frac{1}{(1-D)²}(x+1)$

= e^x (1- D)^-2(x+1)

= e^x (1 + 2D + 3D² + •••••• upto ∞)(x+1)

= e^x{(x+1) + 2(1) + 3(0) ••••• upto ∞}

= e^x(x+3)

= (x+3)e^x

So the general solution is y = C.F. + P.I.

That means general solution is

y = (c₁+c₂x)e^2x + (x+3)e^x where c_1, and c₂ are arbitrary constants.