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How to find the homogeneous and particular solutions for a second-order ordinary ODE, but with a polynomial particular solution
You’ve learned how to solve a second-order ODE that has both a homogeneous and a particular solution, for the cases of trigonometry in the right-hand side, exponential e terms in the right-hand side, and hyperbolic trigonometry terms in the right-hand side.
Let’s cover the final “case”, when a polynomial is present on the right-hand side.
Example problem: Find the solution of \(y” – 10y’ + 25y = 30x + 3\)
Let \(y = {e^{mx}}\), find its derivatives, and substitute into the ODE to get the following polynomial:
\[{m^2} – 10m + 25 = 0\]
You will find repeated real roots, m = 5
To find the particular solution, you need to make another polynomial, consisting of unknown variables A, B, etc.
This polynomial only has a linear term and a constant, but if you had quadratic or cubic terms you’d need to add them in too with extra variables.
Assume the particular solution is in the form \({y_P} = Ax + B\).
Then, \({y_P}’ = A\) and \({y_P}” = 0\).
Substitute these into the original ODE.
\[ – 10A + 25Ax + 25B = 30x + 3\]
Equate the power of x to the 1 terms, and equate the power of x to the 0 terms.
\[25Ax = 30x\]
\[ – 10A + 25B = 3\]
Use these to solve for A and B, which will be A = 6/5 and B = 3/5
Now combine the homogeneous solution and the particular solutions together.
\[y = {c_1}{e^{5x}} + x{c_2}{e^{5x}} + \frac{6}{5}x + \frac{3}{5}\]
How to solve a second-order ordinary differential equation using MATLAB and double checking your answer
If you have the program MATLAB, you can easily use its symbolic tools to get the exact solution to the problem, feel free to use our following script and modify it for the problem you’re trying to solve.
It’s extremely handy for double-checking your answers before handing them in for an assignment. You can even make your own problems and solve them if you want extra practice solving second-order ODEs.
%bai-gaming.com/math-guides
syms y(x) x
eqn = diff(y,x,2) - 10*diff(y,x,1) + 25*y == 30*x + 3;
S = dsolve(eqn)Practice how to find the particular solution for more complicated ODEs!
Next, try using the Laplace Transform to solve second-order ODES!