Answer to Question #122583 in Differential Equations for jse

Question

1. Find the general solution of the given second-order differential equation.
3y'' + y' = 0

y(x) = _____

2. Solve the given initial-value problem.
y''' + 14y'' + 49y' = 0, y(0) = 0, y'(0) = 1, y''(0) = −8

y(x) =____

Answer 1

$3 y^{{\prime}{\prime}}+y^{\prime}=0 \\[1 em] \text{A second order linear homogeneous } \\[1 em] \text{The geveral solution is} \\[1 em] \begin{array}{l} y(x)=y_{h} \\[1 em] 3 y^{{\prime}{\prime}}+y^{\prime}=0 \rightarrow 3r^{2}+r=0 \\[1 em] \therefore r(3 r+1)=0 \rightarrow r=0, r=-\frac{1}{3} \\[1 em] \therefore y(x)=c_{1} e^{0}+c_{2} e^{-\frac{1}{3} x} \\[1 em] (1) \implies y(x)=c_{1}+c_{2} e^{-\frac{1}{3} x} \\[1 em] \end{array}\\[1 em] y^{\prime \prime \prime}+14 y^{\prime \prime}+49 y^{\prime}=0 \\[1 em] \text { Alinear homogeneous solution }\\[1 em] \text { The general solution }\\[1 em] y(x)=y_{h} \\[1 em] r^{3}+14 r^{2}+49 r=0\\[1 em] \therefore r\left(r^{2}+14 r+49\right)=0 \\[1 em] \therefore r(r+7)^{2}=0 \quad \Rightarrow r=0, r=-7, r=-7 \\[1 em] \therefore y(x)=c_{1} e^{0}+c_{2} e^{-7 x}+c_{3} x e^{-7 x} \\[1 em] y(x)=c_{1}+c_{2} e^{-7 x}+c_{3} x e^{-7 x} \\[1 em] y(0)=0 \quad \longrightarrow \quad c_{1}+c_{2}=0~~~~(1) \\[1 em] y^{\prime}(x)=-7 c_{2} e^{-7 x}+c_{3}\left(-7 x e^{-7 x}+e^{-7 x}\right) \\[1 em] y^{\prime}(0)=1 \\[1 em] \therefore \quad 1=-7 c_{2}+c_{3}(0+1) \rightarrow-7_{2}+c_{3}=1 ~~~~(2) \\[1 em] y^{\prime \prime}(x)=49 c_{2} e^{-7 x}+c_{3} \left(49 x e^{-7 x}-7 e^{-7 x}-7 e^{-7 x}\right) \\[1 em] y^{\prime \prime}(x)=49 c_{2} e^{-7 x}+c_{3}\left(49 x e^{-7 x}-14 e^{-7 x}\right) \\[1 em] y^{\prime \prime}(0)=-8 \\[1 em] \therefore \quad-8=49 c_{2}+14 C_{3}~~~~(3) \\[1 em] \text { SoLv } (1),(2),(3) \\[1 em] \therefore c_{1}=\frac{6}{49}, c_{2}=-\frac{6}{49}, c_{3}=\frac{1}{7} \\[1 em] y(x)=\frac{6}{49}-\frac{6}{49} e^{-7 x}+\frac{1}{7} x e^{-7 x} \\[1 em] y(x)=\frac{1}{49} e^{-7 x}\left(6 e^{7 x}+7 x-6\right)$