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subject: Differential Equation Calculus [print this page]

An equation with differentials is called a differential equation. Each differential will have an order and degree. Both order and degree of a differential equation should be in natural numbers. Example of a differential equation is dy / dx + 2y = 5.

Here this differential equation of first order and first degree. The order of the differential equation is the highest order of the derivative in the given differential equation. The power of the highest order derivative is called its degree. Example, In the given differential equation, [d2y] / [d^2x]/[dy^2] + 3[dy/dx ]2 + 4y = 20. The highest order 2 and its degree is 1. Because the power of the highest order of the differential is 1.

We have many methods to find the solution of the differential equation. One among them is variable separable method. To apply this method, the variables should align in such a way that one of the variable is on the left and the other is one the right.

Example Problems on Differential Equation:

Ex 1: Solve the differential equation: y dy x dx = 0.

Sol: Given: ydy xdx = 0.

= y dx = x dx.

Here the variables are separated. So, we can obtain the solution by Integrating them.

therefore int y dy = int x dx = y^2/ 2 = x^2/ 2 + c.

= y^2 = x2 + k, where k = 2c.

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This is the solution.

Ex 2: Form the differential equation of y = [ax2 + bx, where a and b are arbitrary constants.

Sol: Given: y = [ax2 + bx - (1)

Now in (1) we need to eliminate a and b

therefore (1) = [dy] / [dx] = 2ax + b - (2)

Also, [d^2y] / [dx^2] = 2a - (3)

therefore By using (3) in (2), we get

[dy] / [dx] = [x d^2y] / [dx^2] + b

= b = [dy] / [dx] ** [xd^2y] / [dx^2] - (4)

therefore (1) = y = [x^2] / 2 [d^2y] / [dx^2] + [[dy/dx] ** [xd^2y]/ [dx^2]] x.

= y = [x^2] / 2 [d^2y] / [dx^2] + x dy / [dx] ** x^2[d^2y] / [dx^2]

= x^2/2 [d^2y] / [dx^2] ** x dy / [dx] + y = 0.

= x2 [d^2y] / [dx^2] ** 2x [dy] / [dx] + 2y = 0.

Ex 3: Solve: [2y 1] dx [2x + 3] dy = 0.

Sol: (2x + 3) dy = (2y 1) dx.

= 1/ (2y **1)] dy = [1/ (2x + 3)] dx.

int [[dx] / [2x + 3]] = int [[dy] / [2y**1]] + K.

= 1/2 ln (2x + 3) = 1/2 ln ( 2y 1) + K

= 1/2 ln (2x + 3) / (2y ** 1) = K

= ln [2x + 3] / (2y **1) = C! = log c

= 2x + 3] / [2y**1] = c.

[Here Let C! = log c]

Practice Problems on Differential Equation:

1. Solve: sqrt [1 ** y^2] dx + sqrt (1 ** x^2) dy = 0.

[Ans: Sin-1(x) + Sin-1(y) + c]

2. Form the differential equation of xy = a2.

[Ans: x dy / dx + y = 0]

by: mathqa

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