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

Answer:The value of k is 6Step-by-step explanation:we need to find the value of k Given : - [tex]y=e^{2x}[/tex] is the solution [tex]y''-5y'+ky=0[/tex][tex]y=e^{2x}[/tex]                               ........(1)                  differentiate  [tex]y=e^{2x}[/tex] with respect to 'x' [tex]\frac{dy}{dx}=\frac{d}{dx}e^{2x}[/tex]Since, [tex]\frac{d}{dx}e^{x} =e^{x}\frac{d}{dx}(x)[/tex] [tex]\frac{dy}{dx}=e^{2x}\frac{d}{dx}(2x)[/tex] [tex]\frac{dy}{dx}=e^{2x}\times 2[/tex][tex]\frac{dy}{dx}=2e^{2x}[/tex]so, [tex]y'=2e^{2x}[/tex]                     ..........(2)Again differentiation above with respect to 'x'[tex]\frac{d}{dx}\frac{dy}{dx}=\frac{d}{dx}2e^{2x}[/tex][tex]\frac{d^{2}y}{dx^{2}}=2e^{2x}\frac{d}{dx}(2x)[/tex][tex]\frac{d^{2}y}{dx^{2}}=2e^{2x}\times 2[/tex][tex]\frac{d^{2}y}{dx^{2}}=4e^{2x}[/tex]so, [tex]y''=4e^{2x}[/tex]                         ........(3)Now, put the value of [tex]y\ ,y' \ \text{and} \ y''[/tex] in [tex]y''-5y'+ky=0[/tex][tex]4e^{2x}-5(2e^{2x})+(e^{2x})k=0[/tex][tex]4e^{2x}-10e^{2x}+e^{2x}k=0[/tex][tex]-6e^{2x}+e^{2x}k=0[/tex]add both the sides by [tex]6e^{2x}[/tex][tex]e^{2x}k=6e^{2x}[/tex]Cancel out the same terms from left and right sides[tex]k=6[/tex]Hence, the value of k is 6