Answer to Question #221506 in Differential Equations for Yey

Question

Solve the equation. y′′−6y′+9y=0 ;  y(0)=2,y′(0)=−4

a. using differential equation method

b. using Laplace transformation method

Answer 1

a. Characteristic equation:

${k^2} - 6k + 9 = 0$

${(k - 3)^2} = 0$

${k_1} = {k_2} = 3$

Then

$y = {C_1}{e^{3x}} + x{C_2}{e^{3x}} \Rightarrow y' = 3{C_1}{e^{3x}} + {C_2}{e^{3x}} + 3x{C_2}{e^{3x}}$

$y(0) = 2,\,y'(0) = - 4 \Rightarrow \left\{ {\begin{matrix} {{C_1} = 2}\\ {3{C_1} + {C_2} = - 4} \end{matrix}} \right. \Rightarrow {C_1} = 2,\,{C_2} = - 10$

Then

$y = 2{e^{3x}} - 10x{e^{3x}}$

Answer: $y = 2{e^{3x}} - 10x{e^{3x}}$

b. Moving from originals to images:

$y \to Y$

$y' \to pY - y(0) = pY - 2$

$y'' \to {p^2}Y - py(0) - y'(0) = {p^2}Y - 2p + 4$

Then

${p^2}Y - 2p + 4 - 6\left( {pY - 2} \right) + 9Y = 0$

$Y\left( {{p^2} - 6p + 9} \right) = 2p - 4 - 12$

$Y = \frac{{2p - 16}}{{{p^2} - 6p + 9}} = \frac{{2p - 16}}{{{{(p - 3)}^2}}} = \frac{2}{{p - 3}} - \frac{{10}}{{{{(p - 3)}^2}}}$

$Y \leftarrow y$

$\frac{1}{{p - 3}} \leftarrow {e^{3x}}$

$\frac{1}{{{{\left( {p - 3} \right)}^2}}} \leftarrow x{e^{3x}}$

Then

$y = 2{e^{3x}} - 10x{e^{3x}}$

Answer: $y = 2{e^{3x}} - 10x{e^{3x}}$