91Ó°ÊÓ

The given expressions are$y^2+2y+10y=\mathrm{δ}\left(t-t_0\right)$.

A transformation of a function f(x) into the function g(t) that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

The inverse Laplace transform of a functionF(s) is the piecewise-continuous and exponentially-restricted real functionf(t)

Solve the equation

$y^{\text{'}}+2y\text{'}+10y=\mathrm{δ}\left(t{-}_0\right)$

By taking transforms to the given differential equation,

$L\left(y^n\right)+2·L\left(y\right)+10-L\left(y\right)=L\left[\mathrm{δ}\left(t-t_0\right)\right]$

Put the value of

$\left(p^{2}y-p{y}_0-y_0\right)+2\left(pY-y_0\right)+10e^{ey}$

Put $y_0=y_0=0$in the above equation

By equation (2)

$y=\frac{e^{-\left(p-b\right)·}\sin 3\left(t-b_0\right)}{3}if>t_0$

And$y=0;1<a$;

Whereby $a-t_0\left(by\left[1\right]\right)$

And;

$g\left(t-a\right)=\left(t-a\right)·e^{-\left(t-a\right)}ift>t_0>0=0ift<t_0$

By equation (2)

$g\left(t-a\right)=\left(t-a\right)-e^{-\left(t-a\right)}ift<t_0=0ift<t_0$

Which is the general solution of the given differential equation.