# Solving 2nd Order Linear DE with Constant Coefficients

• Apteronotus
In summary, when solving a 2nd order Linear DE with constant coefficients, we look for solutions of the form y=e^{rt} and the solution is given by y(t)=c_1 e^{r_1 t}+c_2 e^{r_2 t}. This is the general solution because it can be proven for the first-order case that there are no other solutions.
Apteronotus
Hi,

When solving a 2nd order Linear DE with constant coefficients ($ay''+by'+cy=0$) we are told to look for solutions of the form $y=e^{rt}$ and then the solution (if we have 2 distinct roots of the characteristic) is given by
$y(t)=c_1 e^{r_1 t}+c_2 e^{r_2 t}$

This is clearly a solution, but how do we know there are no other solutions?
That is, how do we know this is the general solution?

Hi Apteronotus!
Apteronotus said:
… how do we know there are no other solutions?

It's easy to prove for the first-order case …

if y' - ry = 0, put y = zert, then (z' + rz)ert = rzert

so ert = 0 (which is impossible),

or z' + rz = rz, ie z' = 0, ie z is constant

and now try (y' - ry)(y' - sy) = 0, using the same trick twice

## 1. What is a 2nd Order Linear Differential Equation?

A 2nd Order Linear Differential Equation is an equation that involves a second derivative of a function, along with the function itself and possibly a first derivative. It is typically written in the form of: y'' + p(x)y' + q(x)y = g(x), where p(x) and q(x) are continuous functions and g(x) is a known function.

## 2. What are the steps for solving a 2nd Order Linear Differential Equation with Constant Coefficients?

The steps for solving a 2nd Order Linear Differential Equation with Constant Coefficients are as follows:

• Step 1: Find the characteristic equation by replacing y'' with r^2, y' with r, and y with 1.
• Step 2: Solve the characteristic equation to find the roots, r1 and r2.
• Step 3: Determine the form of the solution based on the values of r1 and r2. If both roots are real and distinct, the form is y = c1e^r1x + c2e^r2x. If the roots are complex, the form is y = e^ax(c1cos(bx) + c2sin(bx)), where a and b are the real and imaginary parts of the complex root.
• Step 4: Use initial conditions to find the values of c1 and c2.
• Step 5: Substitute the values of c1 and c2 into the solution to get the final solution.

## 3. Can a 2nd Order Linear Differential Equation have non-constant coefficients?

Yes, a 2nd Order Linear Differential Equation can have non-constant coefficients. In this case, it is called a non-homogeneous equation and the solution involves both a complementary function (the solution to the associated homogeneous equation with all coefficients equal to zero) and a particular integral (a function that satisfies the non-homogeneous equation).

## 4. What are the boundary conditions for a 2nd Order Linear Differential Equation?

The boundary conditions for a 2nd Order Linear Differential Equation are the initial values or constraints given for the function y, its first derivative y', or its second derivative y''. These values are typically given at specific points or over an interval of the independent variable.

## 5. How can I check if my solution to a 2nd Order Linear Differential Equation is correct?

To check if your solution to a 2nd Order Linear Differential Equation is correct, you can plug it back into the original equation and see if it satisfies the equation. You can also check if it satisfies the initial conditions given in the problem. Additionally, you can take the derivatives of your solution and see if they match with the derivatives in the original equation. If all of these checks are true, then your solution is correct.

Replies
3
Views
1K
Replies
2
Views
689
Replies
7
Views
2K
Replies
2
Views
2K
Replies
3
Views
2K
Replies
2
Views
2K
Replies
1
Views
655
Replies
2
Views
2K
Replies
1
Views
2K
Replies
1
Views
2K