Question

for what value of k, if any, is y=e2x ke−3x a solution to the differential equation 4y−y′′=10e−3x ?

Answer 1

Using substitution, it is found that k = 2 satisfies the differential equation.

The solution is:

$y = e^{2x} + ke^{-3x}$

The differential equation is:

$4y - y^{\prime\prime} = 10e^{-3x}$

We replace y and it's derivatives in the solution, then:

$y^{\prime}(x) = 2e^{2x} - 3ke^{-3x}$

$y^{\prime\prime}(x) = 4e^{2x} + 9ke^{-3x}$

Then:

$4y - y^{\prime\prime} = 10e^{-3x}$

$4(e^{2x} + ke^{-3x}) - (4e^{2x} + 9ke^{-3x}) = 10e^{-3x}$

$4ke^{-3x} - 9ke^{-3x} = 10e^{-3x}$

$-5ke^{-3x} = 10e^{-3x}$

$5k = -10$

$k = -\frac{10}{5}$

$k = -2$

k = 2 satisfies the differential equation.

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