Particular solution of second order differential equation

Hints/Guides on how to solve such differential equations :

$\mathbf{1}$ - Method of Undetermined Coefficients :

Start of by solving the homogenous equation $y''-2y'+y= 0$ by assuming that a solution will be proportional to $e^{\lambda t}$ for some $\lambda$. Substitute in and calculate $\lambda$. Notice the multiplicity of the solution for $\lambda$ and adjust your general solution accordingly.

Then, use the method of undetermined coefficients to find a particular solution of the problem for $y''-2y'+y=e^t$.

The general solution of the initial differential equation, will then be the general solution of the homogenous plus the particular solution you found.

You can find more information and examples about that method, here.

$\mathbf{2}$ - Laplace Transformation :

This is a very fast and straight forward way to approach the problem, but it needs some fluent handling of Laplace Transformation techniques. Note that you can apply the Laplace Transformation without even needing initial conditions, simply stating them as constants.

Start of by applying the Laplace Transformation

$$\mathcal{L}_t\big[f(t)\big](s) = \int_0^\infty f(t)e^{-st}\mathrm{d}t$$

to both sides of the given differential equation :

$$\mathcal{L}_t\big[y'' - 2y' + y'] = \mathcal{L}_t[e^t]$$

$$\Leftrightarrow$$

$$(s-1)^2\big[\mathcal{L}_t[y(t)](s)\big] - (s-2)y(0) - y'(0) = \frac{1}{s-1}$$

$$\Leftrightarrow$$

$$\mathcal{L}_t\big[y(t)\big](s) = \frac{y(0)(s^2-3s+2) + y'(0)(s-1) + 1}{(s-1)^3}$$

$$=$$

$$\mathcal{L}_t\big[y(t)\big](s) = \frac{1}{(s-1)^3} - \frac{y(0)}{(s-1)^2} + \frac{y(0)}{s-1} + \frac{y'(0)}{(s-1)^2} $$

$$\implies$$

$$y(t) = \frac{1}{2}e^t(t^2+2c_1 - 2c_1t + 2c_2t) = \frac{e^tt^2}{2} + c_1e^t - c_1e^tt + c_2e^tt$$

$\mathbf{3}$ - Variation of Parameters :

You must repeat the step of solving the homogenous equation by finding that $\lambda$s mentioned. Then by listing the basis solution as $y_{b_1} = e^t$ and $_{b_2} = e^tt$ you can use variation of parameters to find the final general solution by computing the Wronskian and finding the integrals :

$$v_1(t) = - \int \frac{f(t)y_{b_2}(t)}{W(t)}\mathrm{d}t \quad \text{and} \quad v_2(t) = \int \frac{f(t)y_{b_1}(t)}{W(t)}\mathrm{d}t$$

You can find more information and examples about that method, here.

Tags:

Ordinary Differential Equations

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