Q:

Verify that the function(s) solve the following differential equations (DES): a) y' = -5y; y = 3e-5x b) y' = cos(3x); y = į sin(3x) + 7 c) y' = 2y; y = ce2x , where c is any real number. d) y" + y' – 6y = 0 ; yı = (2x, y2 = (–3x e) y" + 16y = 0; yı = cos(4x), y2 = sin(4x)

Accepted Solution

A:
Answer:In the step-by-step explanation, the verifications are made.Step-by-step explanation:a) [tex]y' = -5y[/tex]This one can be solved by the variable separation method[tex]y' = -5y[/tex][tex]\frac{dy}{dx} = -5y[/tex][tex]\frac{dy}{y} = -5dx[/tex][tex]\int \frac{dy}{y}  = \int {-5} \, dx[/tex][tex]ln y = -5x + C[/tex][tex]e^{ln y} = e^{-5x + C}[/tex][tex]y = Ce^{-5x}[/tex]The value of C is the value of y when x = 0. If [tex]y(0) = 3[/tex], then we have the following solution:[tex]y = 3e^{-5x}[/tex]b) [tex]y' = cos(3x)[/tex]This one can also be solved by the variable separation method[tex]y' = cos(3x)[/tex][tex]\int y' \,dy  = \int {cos(3x)} \, dx[/tex][tex]y = \frac{sin(3x)}{3} + K[/tex]K is also the value of y, when x = 0. So, if [tex]y(0) = 7[/tex], we have the following solution.[tex]y = \frac{sin(3x)}{3} + 7[/tex]c) [tex]y' = 2y[/tex]Another one that can be solved by the variable separation method[tex]y' = 2y[/tex][tex]\frac{dy}{dx} = 2y[/tex][tex]\frac{dy}{y} = 2dx[/tex][tex]\int \frac{dy}{y}  = \int {2} \, dx[/tex][tex]ln y = 2x + C[/tex][tex]e^{ln y} = e^{2x + C}[/tex][tex]y = Ce^{2x}[/tex] C is any real number depending on the initial conditions.d) [tex]y'' + y' - 6y = 0[/tex]Here, the solution depends on the roots of the following equation:[tex]r^{2} + r - 6 = 0[/tex][tex]r = \frac{-1 \pm 5}{2}[/tex][tex]r = -3[/tex] or [tex]r = 2[/tex].So the solution is[tex]y(t) = c_{1}e^{-3t} + c2e^{2t}[/tex]The values of [tex]c_{1}, c_{2}[/tex] depends on the initial conditions.e) [tex]y'' + 16y = 0[/tex]Again, we find the roots of the following equation:[tex]r^{2} + 16 = 0[/tex][tex]r^{2} = -16[/tex][tex]r = \pm 4i[/tex]So we have the following solution[tex]y(t) = c_{1}cos(4t) + c_{2}sin(4t)[/tex]The values of [tex]c_{1}, c_{2}[/tex] depends on the initial conditions.