Question

1b. Solve the differential equations

y''+ 4y' + 4y = 0, y(0) = 1 and y'(0) = 1

Note that each equation has right hand side zero. So, you don’t have to use the Laplace transform. You just use the characteristic equaiton.​

Answer 1

Answer:

No solutions

Step-by-step explanation:

$y''+4y'+4y=0,\:y(0)=1,\:y'(0)=1\\\\m^2+4m+4=0\\\\(m+2)(m+2)=0\\\\m=-2$

Thus, since we have equal real roots, we use the general solution $y(x)=C_1e^{m_1x}+C_2e^{m_1x}$ and our initial conditions to set up our system of equations:

$y(x)=C_1e^{-2x}+C_2e^{-2x}\\\\y(0)=C_1e^{-2(0)}+C_2e^{-2(0)}\\\\1=C_1+C_2$ <-- First one

$y(x)=C_1e^{-2x}+C_2e^{-2x}\\\\y'(x)=-2C_1e^{-2x}-2C_2e^{-2x}\\\\y'(0)=-2C_1e^{-2(0)}-2C_2e^{-2(0)}\\\\1=-2C_1-2C_2$ <-- Second one

Solve the system

$C_1+C_2=-2C_1-2C_2\\\\3C_1+3C_2=0\\\\3C_1=-3C_2\\\\C_1=-C_2$

$1=C_1+C_2\\\\1=-C_2+C_2\\\\1=0$

Therefore, there are no solutions to the differential equation given the initial conditions.