Question

If y = e2x is a solution to y''- 5y' + ky = 0, what is the value of k?

Answer 1

Answer:

The value of k is 6

Step-by-step explanation:

we need to find the value of k

Given : - $y=e^{2x}$ is the solution $y''-5y'+ky=0$

$y=e^{2x}$                               ........(1)                  

differentiate  $y=e^{2x}$ with respect to 'x'

$\frac{dy}{dx}=\frac{d}{dx}e^{2x}$

Since, $\frac{d}{dx}e^{x} =e^{x}\frac{d}{dx}(x)$

$\frac{dy}{dx}=e^{2x}\frac{d}{dx}(2x)$

$\frac{dy}{dx}=e^{2x}\times 2$

$\frac{dy}{dx}=2e^{2x}$

so, $y'=2e^{2x}$                     ..........(2)

Again differentiation above with respect to 'x'

$\frac{d}{dx}\frac{dy}{dx}=\frac{d}{dx}2e^{2x}$

$\frac{d^{2}y}{dx^{2}}=2e^{2x}\frac{d}{dx}(2x)$

$\frac{d^{2}y}{dx^{2}}=2e^{2x}\times 2$

$\frac{d^{2}y}{dx^{2}}=4e^{2x}$

so, $y''=4e^{2x}$                         ........(3)

Now, put the value of $y\ ,y' \ \text{and} \ y''$ in $y''-5y'+ky=0$

$4e^{2x}-5(2e^{2x})+(e^{2x})k=0$

$4e^{2x}-10e^{2x}+e^{2x}k=0$

$-6e^{2x}+e^{2x}k=0$

add both the sides by $6e^{2x}$

$e^{2x}k=6e^{2x}$

Cancel out the same terms from left and right sides

$k=6$

Hence, the value of k is 6