Differential Equation

A differential equation is an equation of the form \(\frac{\mathrm{d} y}{\mathrm{d} x} = g(x),\;\;\;\; where\;\; y = f(x)\)

General equations involve Dependent and Independent variables, but those equation which involves variables as well as derivative of dependent variable (y) with respect to independent variable (x) are known as Differential Equation.

The solution of a differential equation is a function, that represents a relationship between the variables, independent of derivatives.

Such as:

Given differential equation : \(\frac{\mathrm{d} y}{\mathrm{d} x} = \cos x\)

Solution :Â \(y = \sin x + c\)

The solution of a differential equation is also known as its primitive. In the upcoming discussions, we will find out the solution of first order and first degree differential equations.

Generally, there are three methods to solve first order and first degree differential equation. We will be discussing only solution of differential equations with separation of variables.

### Variable Separable Differential Equations :

The differential equations which are expressed in terms of (x,y) such that, the x-terms and y-terms can be separated to different sides of the equation (including delta terms). Thus each variable separated can be integrated easily to form the solution of differential equation.

The equations can be written as

\(f(x)dx + g(y)dy = 0\)

where \( f(x) \)

In simpler terms all the differential equations in which all the terms involving \( x ~and~ dx \)

Lets Work Out- Example – Solve \( \frac{dy}{dx} = \frac{x^3 + 3}{y^2 + 1}\) Solution – \((y^2 + 1)dy = (x^3 + 3)dx\) Integrating both the sides we get, \(\int (y^2 + 1)dy = \int(x^3 + 3)dx\) \(\frac{y^3}{3} + y = \frac{x^4}{4} + 3x + c\) \(\frac{y^3}{3} + y – \frac{x^4}{4} – 3x = c\) It is the required solution. Example- Arjun is riding his bike at an initial velocity of 10 m/s. Â To reach his home at time he continuously increases his velocity at the rate of \( 10% \) Solution- Let the velocity of Arjun be v at any time t. Then \(\frac{dv}{dt} = \frac{10}{100} Ã— V\) \(\frac{dv}{dt} = \frac{V}{10}\) Separating the variables we get \( \frac{1}{V}dv = \frac{1}{10}dt \) Integrating both the sides we get; \( ln V = \frac{t}{10} + c \) Where c is any arbitrary constant; \( V = e^{\frac{t}{10}} Ã— e^{c} \) \(V = Ce^{\frac{t}{10}} \) We know at \( t = 0, V = 10 m/s\) So the equation becomes V = \( 10e^{\frac{t}{10}}\) Now \( V = 2.718 Ã— 10 \frac{m}{s}\) Substituting this value we get, \( 27.18 = 10 e^{\frac{t}{10}} \) \( \Rightarrow e^{\frac{t}{10}} = 2.718 \) \( \Rightarrow ln e^{\frac{t}{10}} = ln 2.718 \) \( \Rightarrow \frac{t}{10} = 1\) \( \Rightarrow t = 10 seconds \) |

Exact Differential Equations:-

In the equation

\( f(x)dx + g(y)dy = 0 \)

If \( \frac{\partial d f(x)}{\partial dx}\)

Thus the solution is given by

\( \int f(x) dx + \int g(y) dy = c \)

Lets Work Out- Example- Solve the following :- \( (x^3 + 3xy^2 + 25)dx + (y^3 + 3yx^2 + 25)dy = 0 \) Solution- \(\frac{\partial d f(x)}{\partial dx} = 3x^2 + 3y^2\) \(\frac{\partial d g(y)}{\partial dx} = 3x^2 + 3y^2\) Since\( \frac{\partial d f(x)}{\partial dx} = \frac{\partial d g(y)}{\partial dx}\) Its solution is \(\int f(x)dx + \int g(y)dy = c \) â‡’ \(c = \int(x^3 + 3xy^2 + 25)dx + \int(y^3 + 25)dy\) â‡’ \( c = \frac{x^4}{4} + \frac{y^4}{4} + \frac{3}{2}x^2y^2 + 25(x + y)\) â‡’ \( c = x^4 + y^4 + x^2 y^2 +100(x + y)\) |

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