A differential equation $\frac{di}{dt}-0.2i=0$ is applicable over $-10<t<10$. If $i\left(4\right)=10$, then $i\left(-5\right)$ is _______.

Concept:

Linear Differential Equation: Linear differential equations are those in which the dependent variable, its derivatives occur only in the first degree and they are not multiplied together. It is of the form:

$\frac{{\mathrm{d}}^{\mathrm{n}}\mathrm{y}}{\mathrm{d}{\mathrm{x}}^{\mathrm{n}}}+{\mathrm{k}}_1\frac{{\mathrm{d}}^{\mathrm{n}-1}\mathrm{y}}{\mathrm{d}{\mathrm{x}}^{\mathrm{n}-1}}+{\mathrm{k}}_2\frac{{\mathrm{d}}^{\mathrm{n}-2}\mathrm{y}}{\mathrm{d}{\mathrm{x}}^{\mathrm{n}-2}}+\dots +{\mathrm{k}}_{\mathrm{n}}\mathrm{y}=0$

Where, k1, k2, …kn are the constants.

The solution of the equation is given as:

y = C.F

where C.F is the complementary function.

The above linear differential equation in the symbolic form is represented as

(Dⁿ + k1 Dⁿ⁻¹+ k2 Dⁿ⁻²+…+ kn) y = 0

For different roots of the auxiliary equation, the solution (complementary function) of the differential equation is as shown below.Calculation:

$\begin{array}{l}\frac{di}{dt}-0.2i=0\\ \left(D-0.2\right)i\left(t\right)=0\\ D=0.2\\ i\left(t\right)=k.e^{0.2t},-10<t<10\end{array}$

At $t=4$; $10=K.e^{0.8}$

$\begin{array}{l}K=4.493\\ \therefore i\left(-5\right)=4.493\times e^{-1}\\ i\left(-5\right)=1.65\end{array}$