Question

For what value of k, if any, is y=e−2x+ke4x a solution to the differential equation y−y′′4=5e4x ?

Answer 1

-5/3

Step-by-step explanation:

Correct. One way to verify that a function is a solution to a differential equation is to check that the function and its derivatives satisfy the differential equation. The differential equation in this problem involves both y

and y′′

.

y=e^−2x + ke^4x y′ =−2e^−2x + 4ke^4x y′′= 4e−2x+16ke^4x

Then y−y′′4=e^−2x+ke^4x−(4e−2x+16ke^4x)4=−3ke^4x

.

If −3ke4=5e4x

, then −3k=5⇒k=−53

.

Therefore, y=e−2x+ke4x

will be a solution to the differential equation if k=−5/3

.