Question

Solve the initial value problem. y+2y3y=0;y(0)=6,y(0)=2y^{\prime \prime}+2 y^{\prime}-3 y=0 ; y(0)=6, y^{\prime}(0)=-2

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we solve the associated characteristic equation for the given initial value problem.

λ2+2λ3=(λ+3)(λ1)=0λ1=1,λ2=3Because we have real roots the general solution is associated withy=c1ex+c2e3x\begin{gather*} \lambda^2+2\lambda-3=(\lambda+3)(\lambda-1)=0\\ \Rightarrow \lambda_{1}=1, \lambda_{2}=-3\\ \text{Because we have real roots the general solution is associated with}\\ y=c_{1}e^x+c_{2}e^{-3x} \end{gather*}

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