Find the general solution to the inhomogeneous differential equations.

{eq}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y'' + y' -2y = e^{-2t} {/eq}

## Question:

Find the general solution to the inhomogeneous differential equations.

{eq}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y'' + y' -2y = e^{-2t} {/eq}

## Inhomogeneous Differential Equation:

The inhomogeneous linear differential equation of second order is defined by

{eq}\displaystyle \eqalign{ & y'' + by' + cy = U(t). \cr & \left( {{D^2} + bD + c} \right)y = U(t),\,\,\,D = \frac{d}{{dt}}. \cr} {/eq}

The general solution of this differential equation is obtained by

{eq}\displaystyle y(t) = {y_c}(t) + {y_p}(t) {/eq}

where,

{eq}\displaystyle {y_c}(t) = {c_1}{e^{{m_1}t}} + {c_2}{e^{{m_2}t}}\,\,\,{\text{and}}\,\,\,\,\,{y_p}(t) = \frac{1}{{F(D)}}U(t). {/eq}

## Answer and Explanation: 1

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Given that:

{eq}\displaystyle \eqalign{ & y'' + y' - 2y = {e^{ - 2t}} \cr & \left( {{D^2} + D - 2} \right)y = {e^{ - 2t}},\,\,\,D =...

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Chapter 13 / Lesson 6This lesson explores differential calculus. It defines a differential and delves into the many uses of differential equations.