Question

Verify that the indicated family of functions is a solution of the given differential equation. assume an appropriate interval i of definition for each solution. d2y dx2 − 8 dy dx + 16y = 0; y = c1e4x + c2xe4x

Answer 1

Solution:

Given:  $y = c_{1}e^{4x}+c_{2}xe^{4x}$

${y}'' - 8 {y}' + 16y = 0$            ..... (1)

We can rewrite it as:

$y = \left (c_{1}+c_{2}x \right )e^{4x}$

Find the derivatives.

${y}' = \frac{\mathrm{d} y}{\mathrm{d} x}$

$= \left (c_{1}+c_{2}x \right ) 4e^{4x} + 4c_{2}e^{4x}4$

$= \left (4c_{1} + c_{2} + 4 c_{2}x \right ) e^{4x}$

${y}'' = \left (4c_{1} + c_{2} + 4 c_{2}x \right ) 4e^{4x} + 4c_{2}e^{4x}$

$= \left (16c_{1} + 8c_{2} + 16 c_{2}x \right )e^{4x}$

Substitute the values of $y, {y}' and {y}''$ in equation (1)

${y}'' - 8{y}' + 16y = \left (16c_{1} + 8c_{2} + 16 c_{2}x \right )e^{4x} - 8 \left (4c_{1} + c_{2} + 4 c_{2}x \right ) e^{4x} + 16 \left (c_{1}+c_{2} x \right )e^{4x}$

$= \left \{ 16c_{1} + 8c_{2} + 16c_{2}x - 32c_{1} - 8c_{2} -32c_{2}x + 16c_{1} + 16 c_{2}x \right \}e^{4x}$

$= 0$

Hence Proved.