Question

Verify that given function is a solution of the differential equation

у" — у = 0, y₁(t)=eᵗ, y₂(t) = cosht

Answer 1

The functions y_1(t)=e^t and y_2(t)=cosh(t) are both solutions to the differential equation y'' - y = 0, as their second derivatives minus the original functions equal zero.

Step-by-step explanation:

The question involves verifying whether functions y_1(t)=e^t and y_2(t)=\cosh(t) are solutions to the differential equation y'' - y = 0. To confirm this, we will take the first and second derivatives of the given functions and substitute them into the equation.

Verification for e^t:

First derivative: y'_1(t) = d(e^t)/dt = e^t
Second derivative: y''_1(t) = d^2(e^t)/dt^2 = e^t
Substituting into the equation: e^t - e^t = 0, which confirms that e^t satisfies the differential equation.

Verification for \cosh(t):

First derivative: y'_2(t) = d(\cosh(t))/dt = \sinh(t)
Second derivative: y''_2(t) = d^2(\cosh(t))/dt^2 = \cosh(t)
Substituting into the equation: \cosh(t) - \cosh(t) = 0, which verifies that \cosh(t) is also a solution to the differential equation.