To make things simple and easy, we only look at the case as given below:
$$\frac{d^2y}{dx^2}+P\frac{dy}{dx}+qy=f(x)$$
Fact: Where p and q are constant
The complete solution to such an equation can be found by combining two type of solution which is names as particular solution and Complementary Solution:
SOLUTION:
-
The general solution of the homogeneous equation
$$\frac{d^2y}{dx^2}+P\frac{dy}{dx}+qy=0$$
-
Particular solution of the non homogeneous equation
$$\frac{d^2y}{dx^2}+P\frac{dy}{dx}+qy=f(x)$$
Fact: Note that f(x)could be a single function or a sum of two or more functions.
- At the end we will combine both general and particular solution
The homo generous equation are as follows:
$$\frac{d^2y}{dx^2}-y=0$$
Complementary Solution:
$$y=Ae^x+Be^{-x}$$
The non-homo generous equation
$$\frac{d^2y}{dx^2}-y=2x^2-x-3$$
has a particular solution
$$y=-2x^2+x-1$$
So, the complete solution of differential equation
$$\frac{d^2y}{dx^2}-y=2x^2-x-3$$
is
$$y=Ae^x+Be^{-x}-2x^2+x+1$$
Fact: we have combine these two solutions
Checking:
$$y=Ae^x+Be^{-x}-2x^2+x+1$$
1st and 2nd order derivatives are as follows:
-
$$y=Ae^x-Be^{-x}-4x+1$$
-
$$y=Ae^x+Be^{-x}-1$$
So,
$$\frac{d^2y}{dx^2}-y=Ae^x+Be^{-x}-4-(Ae^x+Be^{-x}-2x^2+x-1)$$
$$\frac{d^2y}{dx^2}-y=Ae^x+Be^{-x}-4-Ae^x-Be^{-x}+2x^2-x+1$$
$$\frac{d^2y}{dx^2}-y=2x^2-x-3$$
How do we find a particular solution?
Example
Find particular Solution of Differential Equation:
$$Y’’+-4y’-12y=3e^{4t}$$
Solution: First we make it homogeneous equation
$$Y’’+-4y’-12y=3e^{4t}$$
The factor equation is:
$$r^2-4r-12=0$$
$$(r-6)(r+2)=0$$
$$r^1=6$$
$$r^2=-2$$
Complementary Solution:
$$y_c=Ae^{6t}+Be^{-2t}$$
Particular Solution:
-
Putting into differential equation after taking derivatives:
After solving we have,
$$3e^{4t}=-7Ae^{4t}$$
$$A=-\frac{3}{7}$$
-
Write right side of equation as
$$y_p=3e^{4t}$$
Particular Solution:
$$y_p=\frac{-3}{7}e^{4t}$$