We can solve the second order differential equation of the type
$$\frac{d^2y}{d^2x}+p(x)\frac{dy}{dx}+Q(x)y=f(x)$$
Fact: Where p(x),Q(x) and f(x) are functions of x , by using
Variation of Parameters
Which only works when f(x) is polynomial, exponential, sine, cosine or a linear combination of those.
Method of Undetermined Coefficient
Which is shortest but works on a wider range of functions.
Fact: But here we start our discussion by learning the case where (x)=0,(it is a homogeneous equation).
$$\frac{d^2y}{d^2x}+p(x)\frac{dy}{dx}+Q(x)y=f(x)$$
And also where a function P(x) and Q(x) are constants p and q
Let’s try to solve them:
e to be rescue
We are going to use a special property of derivative of exponential function
Fact: At any point the slope (derivative ) of ex equal of the value ex:
And when we introduce a value “r” like this:
$$f(x)\;=\;e^{rx}$$
we find
The first derivative is
$$f'(x)\;=\;rx$$
-
The second derivative is
$$f"(x)\;=\;r^2e^{rx}$$
Fact: The first and second derivative are both functions of f(x)
This is going to be helpful to us.
Example
$$\frac{d^2y}{dx^2}+\frac{dy}{dx}-6x=0$$
Let
$$y=e^{rx}$$
So, we have:
- $$\frac{dy}{dx}=re^{rx}$$
- $$\frac{d^2y}{dx^2}=r^2e^{rx}$$
Substitute these into the given equation:
$$r^2e^{rx}+re^{rx-6}e^{rx}=0$$
Simplify:
$$e^{rx}(r^2+r-6)=0$$
$$r^2+r-6=0$$
We have reduced the differential equation to an ordinary quadratic equation.
The quadratic equation is given the special name of the characteristic equation.
We can factor this equation:
$$r^2+r-6$$
$$(r-1)(r+3)=0$$
So,
$$r=2,3$$
And so, we have two solution:
$$y=e^2x$$
$$e^{-3x}$$
General Solution:
$$y=Ae^2x+Be^{-3x}$$