Question

How can I solve this differential equation, \[{y}'''-3{y}''+7{y}'-5y=x\], using undetermined coefficients?

Answer 1

Hint: In order to find solution to this problem, we will first find homogeneous solution that is ${{y}_{h}}\left( x \right)$ and a particular solution that is ${{y}_{p}}\left( x \right)$ and then substitute in the following solution form: $y\left( x \right)={{y}_{h}}\left( x \right)+{{y}_{p}}\left( x \right)$

Complete step by step solution:
We have our given differential equation as:
\[{y}'''-3{y}''+7{y}'-5y=x\]
As we know that the General solution is of the form:
$y\left( x \right)={{y}_{h}}\left( x \right)+{{y}_{p}}\left( x \right)$
Where ${{y}_{h}}\left( x \right)$ is the homogenous solution and ${{y}_{p}}\left( x \right)$ is a particular solution.
Therefore, now we have to find homogenous solution and particular solution and then substitute in the final General solution form.

First, we need to solve the homogenous equation of our given equation that is:
\[{y}'''-3{y}''+7{y}'-5y=0\]
Now the characteristics equation of our equation is:
\[{{r}^{3}}-3{{r}^{2}}+7r-5=0\]
where the power of each term corresponds to the power of the derivative in the homogeneous equation.
On solving the above equation in homogenous equation form, we get the roots as:
\[\begin{align}
  & r=1, \\
 & r=1-2i, \\
 & r=1+2i \\
\end{align}\]
With this, we have one real root that is \[r=1\], which will be an exponential term in our solution.
Also, the complex roots \[r=1\pm 2i\] correspond to sine and cosine multiplied by exponentials.
That is,
$1-2i={{e}^{x}}\sin \left( 2x \right)$ and $1+2i={{e}^{x}}\cos \left( 2x \right)$
Therefore, with this we have our homogenous solution as:
${{y}_{h}}\left( x \right)={{c}_{1}}{{e}^{x}}+{{c}_{2}}{{e}^{x}}\cos \left( 2x \right)+{{c}_{3}}{{e}^{x}}\sin \left( 2x \right)$

Now, we need to find a particular solution.
Therefore, we will consider the RHS of the differential equation that is $x$.
$x$ is a first-order polynomial function.
So, we will make a guess that our particular solution is also a first-order, specifically in the form:
\[{{y}_{p}}(x)=\alpha x+\beta \]
where $\alpha $ and $\beta $ are undetermined coefficients.
We need to calculate up to the third-order derivative of this form because we need to substitute into our differential equation.
So on taking derivative we get:
\[y=\alpha x+\beta \]
First-order derivative:
\[{y}'=\alpha \]
Second-order derivative:
\[{y}''=0\]
Third-order derivative:
${y}'''=0$
Now on, substituting into our differential equation, we get:
\[{y}'''-3{y}''+7{y}'-5y=x\]
\[(0)-3(0)+7(\alpha )-5(\alpha x+\beta )=x\]
On simplifying, we get:
\[7\alpha -5\alpha x-5\beta =x\]
Now on rearranging to make terms on LHS and RHS align more obviously, we get:
\[-5\alpha x+(7\alpha -5\beta )=x\]
Therefore, we can see that:
\[-5\alpha =1\to \left( 1 \right)\]
\[~7\alpha -5\beta =0\to \left( 2 \right)\]
From $\left( 1 \right)$ we find \[\alpha =-\dfrac{1}{5}\] and then substituting this $\alpha $ into $\left( 2 \right)$ we find \[~\beta =-\dfrac{7}{25}\].
Hence, our particular solution is:
${{y}_{p}}\left( x \right)=-\dfrac{1}{5}x-\dfrac{7}{25}$
Since our final solution is:
$y\left( x \right)={{y}_{h}}\left( x \right)+{{y}_{p}}\left( x \right)$
So we will substitute our solution in the following solution, so we get:
\[y(x)={{c}_{1}}{{e}^{x}}+{{c}_{2}}{{e}^{x}}cos(2x)+{{c}_{3}}{{e}^{x}}sin(2x)+\left( -\dfrac{1}{5}x-\dfrac{7}{25} \right)\]
On simplifying, we get:
\[y(x)={{c}_{1}}{{e}^{x}}+{{c}_{2}}{{e}^{x}}cos(2x)+{{c}_{3}}{{e}^{x}}sin(2x)-\dfrac{1}{5}x-\dfrac{7}{25}\]
Therefore,
\[y(x)={{c}_{1}}{{e}^{x}}+{{c}_{2}}{{e}^{x}}cos(2x)+{{c}_{3}}{{e}^{x}}sin(2x)-\dfrac{1}{5}x-\dfrac{7}{25}\] is the solution of the problem.

Note: The degree of the differential equation is the power of the highest order derivative, where the original equation is represented in the form of a polynomial equation in derivatives such as \[{y}',{y}'',\text{ }{y}'''\], and so on.