A differential equation is an equation with one or more functions and their derivatives. Differential Equations also called as Partial differential equations if they have partial derivatives. The highest order derivative is the order of differential equation.

## Differential Equation formula

\(\frac{dy}{dt} + p(t)y = g(t)\) |

p(t) & g(t) are the functions which are continuous.

y(t) = \(\frac{\int \mu (t)g(t)dt + c}{\mu (t)}\)

Where \(\mu (t) = e^{\int p(t)d(t)}\)

## Differential Equation formula question

**Question 1: **

**Solution:**

Differential equation using the integrator factor is:

\(\begin{array}{l}\qquad \begin{aligned} \mathbf{e}^{0.196 t} \frac{d v}{d t}+0.196 \mathrm{e}^{0.196 t} v &=9.8 \mathrm{e}^{0.196 t} \\\left(\mathbf{e}^{0.196 t} v\right)^{\prime} &=9.8 \mathrm{e}^{0.196 t} \end{aligned}\end{array}\)Integrating on both the sides,

\(\begin{array}{l} \int\left(\mathbf{e}^{0.196 t} v\right)^{\prime} d t=\int 9.8 \mathbf{e}^{0.196 t} d t \\ \mathbf{e}^{0.196 t} v+k=50 \mathbf{e}^{0.196 t}+c \\ \mathbf{e}^{0.196 t} v=50 \mathbf{e}^{0.196 t}+c-k \\ \mathbf{e}^{0.196 t} v=50 \mathbf{e}^{0.196 t}+c \\ v(t)=50+c \mathbf{e}^{-0.196 t}\end{array}\)Now, v(0) = 48

⇒ v(0) = 50 + ce^{-0.196(0)}

⇒ 48 = 50 + c

⇒ c = -2

Therefore, v(t) = 50 – 2e^{-0.196t}

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