Solve by the method of undermined coefficients y'' - 6y' + 9y = 4ex.

1.	Solve by the method of undermined coefficients y'' - 6y' + 9y = 4ex.
Answer» Given differential equation is y¹¹ - 6y¹ + 9y = 4e^x It's auxiliarly equation is m² - 6m + 9 = 0 ⇒ (m - 3)² = 0 ⇒ m = 3, 3 \(\therefore\) C.F. = (C₁ + C₂x)e^3x Let y₁ = xe^3x, y₂ = e^3x \(\therefore\) w(y₁,y₂) = \(\begin{vmatrix}xe^{3x}&e^{3x}\\e^{3x}(3x + 1)&3e^{3x}\end{vmatrix}\) = e^6x(3x - 3x - 1) = -e^6x ^{\(\therefore\)} u₁ = \(\int\cfrac{\begin{vmatrix}0&e^{3x}\\4e^x&3e^{3x}\end{vmatrix}}{w(y_1y_2)}dx\) = \(\int\frac{-4e^{4x}}{-e^{6x}}dx\) = 4\(\int e^{-2x}dx\) = \(\frac{4e^{-2x}}{-2}\) = -2e^-2x = 2xe^-2x + e^-2x ⁼(2x + 1)e^-2x \(\therefore\) P.I. = u₁y₂ + u₂y₂ = -2e^-2x x xe^3x + (2x + 1)e^x = e^x \(\therefore\) Complete solution is y = C.F. + P.I. = (C₁ + C₂x)e^3x + e^x

Answer»

Given differential equation is

y¹¹ - 6y¹ + 9y = 4e^x

It's auxiliarly equation is

m² - 6m + 9 = 0

⇒ (m - 3)² = 0

⇒ m = 3, 3

\(\therefore\) C.F. = (C₁ + C₂x)e^3x

Let y₁ = xe^3x, y₂ = e^3x

\(\therefore\) w(y₁,y₂) = \(\begin{vmatrix}xe^{3x}&e^{3x}\\e^{3x}(3x + 1)&3e^{3x}\end{vmatrix}\)

= e^6x(3x - 3x - 1) = -e^6x

^{\(\therefore\)} u₁ = \(\int\cfrac{\begin{vmatrix}0&e^{3x}\\4e^x&3e^{3x}\end{vmatrix}}{w(y_1y_2)}dx\)

= \(\int\frac{-4e^{4x}}{-e^{6x}}dx\) = 4\(\int e^{-2x}dx\)

= \(\frac{4e^{-2x}}{-2}\) = -2e^-2x

= 2xe^-2x + e^-2x

⁼(2x + 1)e^-2x

\(\therefore\) P.I. = u₁y₂ + u₂y₂

= -2e^-2x x xe^3x + (2x + 1)e^x

= e^x

\(\therefore\) Complete solution is y = C.F. + P.I.

= (C₁ + C₂x)e^3x + e^x

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