Differential Equations formula

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.


\(\begin{array}{l}\text { Solution: } \\ \frac{d v}{d t}=9.8-0.196 v \\ \frac{d v}{d t}+0.196 v=9.8 \\ \text { Integrating factor }=\mathbf{e}^{\int 0.196 \, d t}=\mathbf{e}^{0.196 t}\end{array}\) \(\begin{array}{l}\text { Differential equation using the integrator factor is: } \\ \qquad \begin{aligned} \mathbf{e}^{0.196 t} \frac{d v}{d t}+0.196 \mathrm{e}^{0.196 t} v &=9.8 \mathrm{e}^{0.196 t} \\\left(\mathbf{e}^{0.196 t} v\right)^{\prime} &=9.8 \mathrm{e}^{0.196 t} \end{aligned}\end{array}\) \(\begin{array}{l}\text { Integrating on both the sides, } \\ \int\left(\mathbf{e}^{0.196 t} v\right)^{\prime} d t=\int 9.8 \mathbf{e}^{0.196 t} d t \\ \mathbf{e}^{0.196 t} v+k=50 \mathbf{e}^{0.196 t}+c \\ \mathbf{e}^{0.196 t} v=50 \mathbf{e}^{0.196 t}+c-k \\ \mathbf{e}^{0.196 t} v=50 \mathbf{e}^{0.196 t}+c \\ v(t)=50+c \mathbf{e}^{-0.196 t}\end{array}\)

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