## Elementary Differential Equations and Boundary Value Problems 9th Edition

The solution of the initial value problem is $y = {t^{ - 2}}\left[ { - t\cos t + \sin t + {{\left( {\frac{\pi }{2}} \right)}^2} - 1} \right]$
Consider the following initial value problem, $\begin{array}{c} ty' + 2y = \sin t\\ y\left( {\frac{\pi }{2}} \right) = 1 \end{array}$ First, find the integrating factor $\mu \left( t \right)$ of the above first order linear differential equation. Since, the integrating factor $\mu \left( t \right)$ of the general first order linear differential equation $\frac{{dy}}{{dt}} + p\left( t \right)y = g\left( t \right)$ is evaluated as, $\mu \left( t \right) = {e^{\int {p\left( t \right)dt} }}$ And the solution of the first order linear differential equation is evaluated as,$y \cdot \mu \left( t \right) = \int {\mu \left( t \right) \cdot g\left( t \right)dt + c}$ Where $c$ is constant of integration. Now, write the given first order linear differential equation in standard form. $t\frac{{dy}}{{dt}} + 2y = \sin t$ Divide both sides in the above equation by $t$, then $\frac{{dy}}{{dt}} + \frac{2}{t}y = \frac{{\sin t}}{t}$ Now, compare the above differential equation with the general first order linear differential equation. Then, $p\left( t \right) = \frac{2}{t},g\left( t \right) = \frac{{\sin t}}{t}$ Therefore, the integrating factor of the first order differential equation is, $\begin{array}{c} \mu \left( t \right) = {e^{\int {p\left( t \right)dt} }}\\ = {e^{\int {\frac{2}{t}dt} }}\\ = {e^{2\ln \left| t \right|}}\\ = {e^{\ln {{\left| t \right|}^2}}}\\ = {t^2} \end{array}$ Since, the solution of the first order differential is evaluated as, $y \cdot \mu \left( t \right) = \int {\mu \left( t \right) \cdot g\left( t \right)dt + c}$ Now, put the value of integrating factor $\mu\left(t\right)$ and ${g\left( t \right) = \frac{{\sin t}}{t}}$. . Then, $\begin{array}{c} y \cdot \mu \left( t \right) = \int {\mu \left( t \right) \cdot g\left( t \right)dt + c} \\ y \cdot {t^2} = \int {{t^2} \cdot \frac{{\sin t}}{t}dt + c} \\ y \cdot {t^2} = \int {t \cdot \sin tdt + c} \\ y \cdot {t^2} = - t \cdot \cos t + \sin t + c \end{array}$ Then, the general solution of the first order differential equation is, $y \cdot {t^2} = - t\cos t + \sin t + c$ Now, apply the initial condition$y\left( {\frac{\pi }{2}} \right) = 1$ Then, $\begin{array}{c} 1 \cdot {\left( {\frac{\pi }{2}} \right)^2} = - \left( {\frac{\pi }{2}} \right)\cos \left( {\frac{\pi }{2}} \right) + \sin \left( {\frac{\pi }{2}} \right) + c\\ {\left( {\frac{\pi }{2}} \right)^2} = - \left( {\frac{\pi }{2}} \right) \cdot 0 + 1 + c\\ {\left( {\frac{\pi }{2}} \right)^2} = 1 + c\\ c = {\left( {\frac{\pi }{2}} \right)^2} - 1 \end{array}$ Now, put the value of the constant $c$ in the general solution of the first order differential equation, $\begin{array}{c} y \cdot {t^2} = - t\cos t + \sin t + c\\ y \cdot {t^2} = - t\cos t + \sin t + {\left( {\frac{\pi }{2}} \right)^2} - 1\\ y = {t^{ - 2}}\left[ { - t\cos t + \sin t + {{\left( {\frac{\pi }{2}} \right)}^2} - 1} \right] \end{array}$ Hence, the required solution of the initial value problem is $y = {t^{ - 2}}\left[ { - t\cos t + \sin t + {{\left( {\frac{\pi }{2}} \right)}^2} - 1} \right]$