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Considering an equation: $$ 4y''-12y'+9y= e^{5t}+e^{3t}$$

Solving the homogeneous equation

$$ y_h= c_1e^{1.5t}+c_2te^{1.5t}$$

Now to find the general solution of the given equation, Can you resolve yourself by the following method? $$ 4y_1''-12y_1'+9y_1= e^{5t}$$ and

$$4y_2''-12y_2'+9y_2= e^{3t}$$ Then $$y_1 =Ae^{5t}$$ and $$y_2 =Be^{3t}$$

Making the first and second derivative of y1 e y2.

and substituting in the initial equation.

$$A=\frac{1}{49}$$ and $$ B = \frac{1}{9}$$

Then the solution geral of equation is: $$ y(t) =c_1e^{1.5t}+c_2te^{1.5t}+\frac{1}{49}e^{5t}+\frac{1}{9}e^{3t}$$

Is this a correct method to solve? Thanks!!

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    $\begingroup$ This is the answer that Wolfram Alphra gives me, so it appears to be correct. I believe this method will only work when the right-hand side is a linear combination of $e^{at}$, $a\in\mathbb R$ though. $\endgroup$
    – Math1000
    Dec 1, 2019 at 19:35
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    $\begingroup$ @Math1000 : With polynomial coefficients. Complex exponential factors are also possible. See "method of undetermined coefficients". $\endgroup$
    – Lutz Lehmann
    Dec 1, 2019 at 19:58
  • $\begingroup$ Have you checked your work by plugging your solution into the original equation? $\endgroup$
    – amd
    Dec 1, 2019 at 20:35

1 Answer 1

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Yes, your solution is correct. The given equation is a second order, linear, nonhomogeneous differential equation with constant coefficients. The solution is given by

$$y_g(t)=y_h(t)+y_p(t)$$

where $y_g(t)$ is the general solution, $y_h(t)$ is the homogeneous solution, and $y_p(t)$ is the particular solution. You first found the homogeneous solution as

$$y_h(t)= c_1e^{1.5t}+c_2te^{1.5t}$$

then to find the particular solution you need to solve

$$ 4y''-12y'+9y= e^{5t}+e^{3t}$$

from which you applied the superposition principle and the method of undetermined coefficients. To find the particular solution, you need to solve

$$4y''-12y'+9y= e^{5t}$$

and

$$4y''-12y'+9y= e^{3t}$$

and then use the superposition principle to add these two solutions together. The particular solution will therefore take the form

$$y_p(t)=y_{p_1}(t)+y_{p_2}(t)$$

The method of undetermined coefficients forms two guesses $$y_{p_1}(t)=Ae^{5t},\quad y_{p_2}(t)=Be^{3t}$$

from which you can relate the coefficients to find

$$A=\frac{1}{49},\quad B=\frac{1}{9}$$

which implies that the particular solution is

$$y_p(t)=y_{p_1}(t)+y_{p_2}(t)=Ae^{5t}+Be^{3t}=\frac{1}{49}e^{5t}+\frac{1}{9}e^{3t}$$

therefore the general solution is given as

$$y_g(t)=y_h(t)+y_p(t)=c_1e^{1.5t}+c_2te^{1.5t}+\frac{1}{49}e^{5t}+\frac{1}{9}e^{3t}$$

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