## Related questions with answers

Three solutions of a second-order linear equation L[y] = g(t) are $\psi_{1}(t)=3 e^{t}+e^{t^{2}}, \psi_{2}(t)=7 e^{t}+e^{t^{2}}$ and $\psi_{3}(t)=5 e^{t}+e^{-t^{3}}+e^{t^{2}}.$ Find the solution of the initial-value problem $L[y]=g ; \quad y(0)=1, \quad y^{\prime}(0)=2.$

Solution

Verified$\begin{align*} \textbf{The general equation [1]:}\\ L[y]&=\dfrac{d^{2}y}{dt^{2}}+p(t)\dfrac{dy}{dt}+q(t)y \dots \text{[1]}\\\\ L[y]&=g(t)\\&=\dfrac{d^{2}y}{dt^{2}}+p(t)\dfrac{dy}{dt}+q(t)y\\\\ \textbf{Three solutions of a second-order linear equation [1] are:}\\ \psi_{1}(t)&=3e^{t}+e^{t^{2}}\\ \psi_{2}(t)&=7e^{t}+e^{t^{2}}\\ \psi_{3}(t)&=5e^{t}+e^{-t^{3}}+e^{t^{2}}\\ \textbf{is a solution corresponding homogenous equation:}\\ \psi_{2}(t)-\psi_{1}(t)&=4e^{t}\\ \psi_{3}(t)-\psi_{2}(t)&=e^{-t^{3}}-2e^{t}\\\\ \Rightarrow y(t)&=4c_{1}e^{t}+c_{2}(e^{-t^{3}}-2e^{t})+e^{t^{2}}\\ \textbf{is the general solution.}\\\\ \end{align*}$

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