Answer to Question #126298 in Differential Equations for jse

Question

Solve the initial value problem

y′′+4y′+4y=0, y(−1)=4, y′(−1)=4

Answer 1

$y''+4y'+4y=0$

initial values:

$y(-1)=4$

$y'(-1)=4$

Step by step solution: $\tfrac{d^2y(x)}{dx^2} + 4 \tfrac{dy(x)}{dx} + 4y(x) = 0$

1-step: Assume a solution will be proportioanl to $e^{\lambda x}$ for some constant $\lambda$ . Substitute $y(x) = e^{\lambda x}$ into the differential equation:

$\tfrac{d^2 (e^{\lambda x})}{dx^2} + 4 \tfrac{d (e^{\lambda x})}{dx} + 4e^{\lambda x} = 0$

2-step: Substiute $\tfrac{d^2 (e^{\lambda x})}{dx^2} = \lambda ^2 e^{\lambda x}$ and $\tfrac{d (e^{\lambda x})}{dx} = \lambda e^{\lambda x}$

$\lambda^2 e^{\lambda x} + 4\lambda e^{\lambda x} + 4e^{\lambda x} = 0$

3-step: Factor out $e^{\lambda x}$

$(\lambda^2+4\lambda+4)e^{\lambda x} = 0$

4-step: Since $e^{\lambda x} {=}\mathllap{/\,} 0$ for any finite $\lambda$ , the zeros must come from the polynomial $\lambda^2+4\lambda+4=0$

5-step: Factor

$(\lambda + 2)^2 = 0$

6-step: Solve for $\lambda$

$\lambda = -2$ or $\lambda = -2$

7-step: The multiplicity of the root $\lambda = -2$ is 2 which gives y₁(x) = C₁ $e^{-2x}$ , y₂(x) = C₂ $e^{-2x} x$ as solutions, where C₁ and C₂ are arbitrary constants.

The general solution is the sum of the above solutions:

y(x) = y₁(x) + y₂(x) = C₁$e^{-2x}$ + C₂$e^{-2x}x$

8-step: Solve for the unknown constants using the given initial conditions

Compute $\tfrac{dy(x)}{dx}:$

$\tfrac{dy(x)}{dx} = \tfrac{d }{dx}$ (C₁$e^{-2x} +$ C₂$e^{-2x}x$ ) = - 2C₁$e^{-2x} +$ C₂$e^{-2x} -$ 2C₂$e^{-2x} x$

9-step: Substitute y(-1) = 4 into y(x) = C₁$e^{-2x} + e^{-2x}x$C₂

$e^{2}$ C₁ - $e^{2}$ C₂ = 4

10-step: Substitute y'(-1) = 4 into $\tfrac{dy(x)}{dx} = -2e^{-2x}$ C₁ + $e^{-2x}$ C₂ - $2e^{-2x}x$ C₂

-2C₁$e^{2} + 3e^{2}$ C₂ = 4

11-step: Solve the system:

C₁ = $\tfrac{16}{e^{2}}$

C₂ = $\tfrac{12}{e^{2}}$

Substitute C₁ = $\tfrac{16}{e^{2}}$ and C₂ = $\tfrac{12}{e^{2}}$ into y(x) = $e^{-2x}$ C₁ + $e^{-2x}x$ C₂

Answer:$y(x) = 4e^{-2(x+1)}(3x+4)$