00:01
In this question we are asked to find a general solution of the given differential equation.
00:04
So let's start to solve this problem.
00:07
Here this is a linear differential equation with constant coefficients.
00:11
Solution of this type of differential equation is given by y is equal to y cf.
00:16
Cf means complementary function plus y pi.
00:19
Pi means particular integral.
00:22
We can write given differential equation as dy over d x minus 2y is equal to x square plus 3.
00:33
Now we substitute d over d x is equal to d.
00:39
After this substitution this differential equation becomes d y minus 2y is equal to x square plus 3.
00:49
Now we take y common d minus 2 y is equal to x square plus 3 fd into y is equal to x square plus 3 fd here fd is equal to d minus 2.
01:07
Now first we will find ycf.
01:11
To find ycf, first we will solve the auxiliary equation for the given differential equation.
01:18
Auxiliary equation for the given differential equation is fm is equal to 0...