Solve the differential equation by variation of parameters. y'' - 9y = 9x/e^3x

Solution:

Given, the differential equation is y'' - 9y = 9x/e^3x.

We have to solve the differential equation by variation of parameters.

The standard form of equation is given by y'' + py’ + qy = X,

Where p, q and X are functions of x.

The complementary function of the equation is given by Yc = C₁Y₁ + C₂Y₂

The complementary function of the given equation is y'' + y = 0

Let y'' = m

Now, m² - 9 = 0

m² = 9

Taking square root,

m = ±3

So, Yc = C₁.e^3x + C₂.e^-3x.

The particular integral is given by Yp = U₁Y₁ + U₂Y₂

Where, U₁ = ∫W₁/W dx

U₂ = ∫W₂/W dx

W = \(\begin{vmatrix}Y_{1} &Y_{2} \\dY_{1} &dY_{2} \\\end{vmatrix}\)

Here, f(x) = 9x/e^3x = 9xe^-3x, Y₁ = e^3x and Y₂ = e^-3x

So, dY₁ = 3e^3x

dY₂’ = -3e^-3x

W = \(\begin{vmatrix}e^{3x} & e^{-3x} \\3e^{3x} & -3e^{-3x} \\\end{vmatrix}\)

= e^3x(-3e^-3x) - e^–3x(3e^3x)

We know, e^3x ×e^-3x = 1

= -3 - 3

W = -6

W₁ = \(\begin{vmatrix}0 &Y_{2} \\f(x) &dY_{2} \\\end{vmatrix}\)

W₁ =\(\begin{vmatrix}0 & 9xe^{-3x} \\ e^{-3x} & -3e^{-3x} \\\end{vmatrix}\)

= 0(-3e^-3x) - 9xe^-3x(e^-3x)

= 0 - 9xe^-6x

W₁ = -9xe^-6x

W₂ = \(\begin{vmatrix}Y_{1} &0 \\dY_{1} &f(x) \\\end{vmatrix}\)

= \(\begin{vmatrix}e^{3x} & 3e^{3x} \\ 0 & 9xe^{-3x} \\\end{vmatrix}\)

= e^3x(9xe^-3x) - 0(3e^3x)

= 9x - 0

W₂ = 9x

\(U_{1} = \int\frac{-9xe^{-6x}}{-6}dx \\ = \frac{9}{6}\int xe^{-6x} dx \\=\frac{3}{2}\int xe^{-6x} dx\)

Let u = x, du = dx

v = (-1/6)e^-6x, dv = e^-6x

\(U_{1}=\frac{3}{2}[\frac{-1}{6}xe^{-6x}-\int\frac{1}{6}e^{-6x}]dx \\=\frac{3}{2}[\frac{-1}{6}xe^{-6x}+\frac{1}{36}e^{-6x}]\\=\frac{-3}{12}xe^{6x}+\frac{3}{72}e^{-6x}\)

U₁ = (-1/4)xe^-6x + (1/36)e^-6x

\(U_{2}=\int \frac{9}{-6}x dx\\=\frac{9}{-6}\int x dx\\=\frac{3}{-2}\int x dx\)

\(U_{2}=\frac{3}{-2}(\frac{x^{2}}{2})\)

U₂ = -3x²/4

Now, Yp = ((-1/4)xe^-6x + (1/36)e^-6x)(e^3x) + (-3x²/4)(e^-3x)

Yp = (-1/4)xe^-3x + (1/24)e^-3x - (3/4)x²e^-3x

We know, the complete solution is Y = Yc + Yp

Y = C₁ e^3x + C₂ e^-3x + (-1/4)xe^-3x + (1/24)e^-3x - (3/4)x²e^-3x

Y = C₁ e^3x + C₂ e^-3x - (1/4)xe^-3x + (1/24)e^-3x - (3/4)x²e^-3x

Therefore, the required solution is Y = C₁ e^3x + C₂ e^-3x - (1/4)xe^-3x + (1/24)e^-3x - (3/4)x²e^-3x.

---

Solve the differential equation by variation of parameters. y'' - 9y = 9x/e^3x

Summary:

The solution of the differential equation y'' - 9y = 9x/e^3x by variation of parameters is Y = C₁ e^3x + C₂ e^-3x - (1/4)xe^-3x + (1/24)e^-3x - (3/4)x²e^-3x.