Question

In each problem 7 through 14, verify that each given function is a solution of the differential equation. Y" - y = 0; y1(t) = et, y2(t) = cosh t.

Answer 1

Answer:

Since L.H.S = R.H.S = 0, for both $y_{1} (t) = e^{t}$ and $y_{2} (t) = cosh(t)$, y₁ and y₂ both satisfy the equation y" - y = 0 and are thus solutions to the differential equation.

Step-by-step explanation:

To check whether the given functions are solutions the given differential equation, we differentiate the functions and then insert it into the given equation.

So y" - y = 0 and  

$y_{1} (t) = e^{t}\\y_{1}' (t) = e^{t}\\ y_{1}" (t) = e^{t}$

Substituting these values of y and y" into the left hand side of the equation, we have

y" - y  

$y_{1}" (t) - y_{1} (t) = e^{t} - e^{t} = 0$

Since L.H.S = R.H.S

So $y_{1} (t) = e^{t}$ is a solution of the differential equation.

When

$y_{2} (t) = cosh(t)\\ y_{2}'(t) = sinh(t) \\y_{2}"(t) = cosh(t)$

Substituting y and y" into the left hand side of the equation, we have

y" - y

$y_{2}"(t) - y_{2} (t) = cosh(t) - cosh(t) = 0$

Since L.H.S = R.H.S

So, $y_{2} (t) = cosh(t)$ is a solution of the differential equation.