Solve by the method of variation of parameters y'' - 4y' + 4y = x-1 e

1.	Solve by the method of variation of parameters y'' - 4y' + 4y = x-1 e2x.
Answer» Given differential equation is y'' - 4y' + 4y = \(\frac{e^{2x}}x\) It's auxiliarly equation is m² - 4m + 4 = 0 ⇒ (m - 2)² = 0 ⇒ m - 2 = 0 ⇒ m = 2, 2 \(\therefore\) w(y₁, y₂) = \(\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}\) = \(\begin{vmatrix}e^{2x}&xe^{2x}\\2e^{2x}&(2x+1)e^{2x}\end{vmatrix}\) = (2x + 1 - 2x)e^4x = e^4x _u1 = \(\int\cfrac{\begin{vmatrix}0&xe^{2x}\\ \frac{e^{2x}}x&(2x+1)e^{2x}\end{vmatrix}}{w(y_1y_2)}dx\) \(=\int\frac{-e^{4x}}{e^{4x}}dx=-x\) u₂ = \(\int\cfrac{\begin{vmatrix}e^{2x}&0\\ 2e^{2x}&\frac{e^{2x}}x\end{vmatrix}}{w(y_1y_2)}dx\) = \(\int\frac{e^{4x}}{x.e^{4x}}dx\) = \(\int\frac1{x}dx=log x\) \(\therefore\) P.I. = u₁y₁ + u₂y₂ = -xe^2x + xe^2xlog x \(\therefore\) Complete solution is y = C.F. + P.I. = (C₁ + C₂x)e^2x - xe^2x + xe^2x log x

Solve by the method of variation of parameters y'' - 4y' + 4y = x-1 e2x.

Answer»

Given differential equation is

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

It's auxiliarly equation is

m² - 4m + 4 = 0

⇒ (m - 2)² = 0

⇒ m - 2 = 0

⇒ m = 2, 2

\(\therefore\) w(y₁, y₂) = \(\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}\) = \(\begin{vmatrix}e^{2x}&xe^{2x}\\2e^{2x}&(2x+1)e^{2x}\end{vmatrix}\)

= (2x + 1 - 2x)e^4x = e^4x

_u1 = \(\int\cfrac{\begin{vmatrix}0&xe^{2x}\\ \frac{e^{2x}}x&(2x+1)e^{2x}\end{vmatrix}}{w(y_1y_2)}dx\)

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

u₂ = \(\int\cfrac{\begin{vmatrix}e^{2x}&0\\ 2e^{2x}&\frac{e^{2x}}x\end{vmatrix}}{w(y_1y_2)}dx\)

= \(\int\frac{e^{4x}}{x.e^{4x}}dx\) = \(\int\frac1{x}dx=log x\)

\(\therefore\) P.I. = u₁y₁ + u₂y₂ = -xe^2x + xe^2xlog x

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

= (C₁ + C₂x)e^2x - xe^2x + xe^2x log x

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