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bezi_cat's question at Yahoo! Answers (Unknown initial condition)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello bezi_cat,

We have the equation: $$y''-25y=e^{-t}\quad(1)$$ The roots of the characteristic equation $\lambda^2-25=0$ are $\lambda=\pm 5$ so, the general solution of the homogeneous is $y_h(t)=C_1e^{5t}+C_2e^{-5t}$. According to a well-known theorem, a particular solution for $(1)$ has the form $y_p(t)=Ke^{-t}$. Substituting in $(1)$ we get $(K-25K)e^{-t}=e^{-t}$ so, $K=-1/24$ that is, the general solution of $(1)$ is: $$y(t)=-\dfrac{1}{24}e^{-t}+C_1e^{5t}+C_2e^{-5t}$$ If $t\to +\infty$ then, $e^{-t}\to 0$, $e^{-5t}\to 0$ and $e^{5t}\to+\infty$. This means that $\lim_{t\to +\infty}y(t)=0$ if and only if $C_1=0$ so, the solution of the IVP has the form $$y(t)=-\dfrac{1}{24}e^{-t}+C_2e^{-5t}$$ The condition $y(0)=1$ implies $\frac{-1}{24}+C_2=1$ that is, $C_2=\frac{25}{24}$. Now, we only need to compute $y'(0)$ where $y(t)=-\frac{1}{24}e^{-t}+\frac{25}{24}e^{-5t}$.