Differential Equations formula

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:

\(\frac{dv}{dt}\) = 9.8 – 0.196v, v(0) = 48 , solve this equation.

Solution:

\(\begin{array}{l} \frac{d v}{d t}=9.8-0.196 v \\ \frac{d v}{d t}+0.196 v=9.8 \\ \text { Integrating factor }=\mathbf{e}^{\int 0.196 \, d t}=\mathbf{e}^{0.196 t}\end{array}\)

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

To explore more formulas on this and other mathematical topics, Register at BYJU’S.

Leave a Comment

Your email address will not be published. Required fields are marked *