Home»Math Guides»Practice Test – Setting Up Particular Solutions for Solving Second-order ODEs

# Practice Test! Learning how to set up particular solution forms when solving second-order ODEs! (Method of Undetermined Coefficients Examples)

If you haven’t already, please review example 1, example 2, example 3, and example 4 for solving second-order ODEs using a homogeneous and particular solution.

The really tricky part is assuming a particular solution form, it’s very easy to mess it up and we’ve seen many students mess up the entire question.

Let’s practice many second-order ODEs, but instead of actually solving them completely, let’s instead just answer what we would assume the particular solution form to be!

This is sometimes asked as an examination question, “Explain the particular solution form you would use for solving the second-order ODE, but don’t actually solve it!”. It really means you need to know your theory well!

Remember to always find the homogeneous solution first on your own! If the particular solution has terms in the homogeneous solution, you need to multiply by x.

**Example problem**: What particular solution form would you use for \(y” + 3y = – 48{x^2}{e^{3x}}\) ?

Now we have a product of a polynomial and an exponential term, so that’s the form of our particular solution, remember to have the highest degree term and also include all lower degree terms too. Also be careful and keep the e’s exponent to be correct.

\[{y_p} = (A{x^2} + Bx + C){e^{3x}}\]

**Example problem**: What particular solution form would you use for \(4y” – 4y’ – 3y = \cos (2x)\) ?

Remember to always include both sine plus cosine, even if you only have one in the actual ODE! Also remember to match the correct frequency.

\[{y_p} = A\cos \left( {2x} \right) + B\sin \left( {2x} \right)\]

**Example problem**: What particular solution form would you use for \(y” + 2y’ = 2x + 5 – {e^{ – 2x}}\) ?

If you find the homogeneous solution, you get roots of 0 and -2. \({e^0}\) conflicts with the polynomial, so that means you need to need to **multiply the whole polynomial by x**. As well, \({e^{ – 2}}\) conflicts, so you need to **multiply the e in the particular solution by x**.

\[{y_p} = A{x^2} + Bx + Cx{e^{ – 2x}}\]

**Example problem**: What particular solution form would you use for \(y” – y’ + \frac{y}{4} = 3 + {e^{\frac{x}{2}}}\) ?

The trick here is that the homogeneous solution has two real repeated roots at 1/2, conflicting with the particular solution, so you need to **multiply the e term in the particular solution by x twice**.

\[{y_p} = A + B{x^2}{e^{\frac{x}{2}}}\]

**Example problem**: What particular solution form would you use for \(y” + 4y = 3\sin (2x)\) ?

The trick is that the roots of the homogeneous solution are complex, leading to cosine and sine terms of 2x So you need to multiply both the sine and cosine terms in the particular solution by x.

\[{y_p} = Ax\sin (2x) + Bx\cos (2x)\]

**Example problem**: What particular solution form would you use for \(y” – 4y = \left( {{x^2} – 3} \right)\sin \left( {2x} \right)\) ?

Product of a polynomial and trigonometry, don’t forget we need both sine and cosine.

\[{y_p} = \left( {A{x^2} + Bx + C} \right)\cos \left( {2x} \right) + \left( {D{x^2} + Ex + F} \right)\sin \left( {2x} \right)\]

**Example problem**: What particular solution form would you use for \(y” + y = 2x\sin \left( x \right)\) ?

The trick is you get cosine and sine in the homogeneous solution due to complex roots. So you need to multiply both the sine and cosine terms by x.

\[{y_p} = \left( {A{x^2} + Bx} \right)\cos \left( x \right) + \left( {C{x^2} + Dx} \right)\sin \left( {2x} \right)\]

**Example problem**: What particular solution form would you use for \(y” – 5y’ = 2{x^3} – 4{x^2} – x + 6\) ?

The trick is that the homogeneous solution has roots of 5 and 0. The 0 is the problem because \({e^0}\) is a constant, and a constant is present in our polynomial for our particular solution. **That means we need to multiply the entire polynomial by x**.

\[{y_p} = A{x^4} + B{x^3} + C{x^2} + Dx\]

**Example problem**: What particular solution form would you use for \(y” – 2y’ + 5y = {e^x}\cos \left( {2x} \right)\) ?

This one, like the others, has a trick. Find the homogeneous solution, you actually have three roots, 0, and two complex roots. The homogeneous solution is \({y_h} = {e^x}\left[ {{c_1}\cos \left( {2x} \right) + {c_2}\sin \left( {2x} \right)} \right]\)

Now, our particular solution is of the form e to the x times sine and cosine, **but since that’s present you need to multiply them by x.**

Be careful, we saw many students who said “e is present and cosine/sine is present, I need to multiply by x twice!”, but that’s wrong!

If you had e OR sine/cosine, you wouldn’t need to multiply by x at all because e **OR** sine/cosine is not the same as e **times** sine/cosine. **Since the homogeneous solution has e times sine/cosine, you multiply by x once. **

\[{y_p} = Ax{e^x}\cos \left( {2x} \right) + Bx{e^x}\sin \left( {2x} \right)\]

Last, but not least, **Example problem**: What particular solution form would you use for \(y” – 2y’ – 6y = {x^4}{e^{6x}}\cos (4x)\) ?

The homogeneous solution has roots of 0, and real numbers, but it’s okay because the e to the 0 is not multiplied by trig or e to the 6, so you don’t actually have particular solution terms appearing in the homogeneous solution because everything is multiplied; if the right-hand side was added instead of multiplied together, then the terms would technically be appearing.

\[{y_p} = (A{x^4} + B{x^3} + C{x^2} + D{x^1} + E){e^{6x}}\sin (4x) + (F{x^4} + G{x^3} + H{x^2} + I{x^1} + J){e^{6x}}\cos (4x)\]

Next, try using the Laplace Transform to solve second-order ODES!