Question

Y = ae^x + be^-x find differential equation

Answer 1

Pre - requiste to understand my answer
1) Differentiation of exponential functions.

Aim to produce differential equation
1) It should not have arbritrary constants like a,b,c here.
Final differential equation is y'' = y.
(independent of a,b,c).

Answer 2

Answer:

$\frac{ {d}^{2} y}{d {x}^{2} } - y = 0 \\ \:$

Step-by-step explanation:

To find the differential equation of

$y = a {e}^{x} + b{e}^{ - x}...eq1$

take first order derivative of y,with respect to x

$\frac{dy}{dx} =a {e}^{x} - b{e}^{ - x} ...eq2$

Now take second derivative,since equation has two arbitrary constant

$\frac{ {d}^{2} y}{d {x}^{2} } = a{e}^{x} + b{e}^{ - x}...eq3$

put value from eq3 to eq1

$\frac{ {d}^{2} y}{d {x}^{2} } = y \\ \\ \frac{ {d}^{2} y}{d {x}^{2} } - y = 0$

is the differential equation.

Hope it helps you.