Stuck in a constant coefficient case question in linear algebra

Question

In advanced engineering mathematics 7th edition book, I stuck in a question, if the saying is right, I tried so hard and got so far :D. This is a linear second order differential equation with constant coefficient case problem. So the questios is to solve the ODE $$y''-y'+4y = 0$$ with the initial conditions $y(-2) = 1$ and $y'(-2) = 3$.

I got to here (which is I am not truly sure if it is correct): $$ e=c_1 \cos(\sqrt{15}) +c_2 \sin(\sqrt{15})\\ \frac{6e}{\cos(\sqrt{15}) +\sin(\sqrt{15})} = c_1 (1+\sqrt{15})+c_2 (1-\sqrt{15}) $$ where $c_1, c_2$ are constants for this equation.

how can I abbrevate this equation and find $c_1, c_2$? Thanks.

Answer 1

After reading TrialandError's post, I discovered that I used the wrong general expression. In TrialandError's post, the roots are not correct. The roots are

$r_\pm = \frac{1 \pm \sqrt{15}}{2}$.

The general expression is

$y(t) = A\exp\Big(\frac{1-i\sqrt{15}}{2}t\Big) + B\exp\Big(\frac{1+i\sqrt{15}}{2}t\Big)$.

Set $\sqrt{15}=r$.

$y(-2) = A\exp(-(1-ir)) + B\exp(-(1+ir))$.

$y'(t) = A\Big(\frac{1-ir}{2}\Big)\exp\Big(\frac{1-ir}{2}t\Big) + B \Big(\frac{1+ir}{2}\Big)\exp\Big(\frac{1-ir}{2}t\Big).$

$y'(-2) = A\Big(\frac{1-ir}{2}\Big)\exp(-(1-ir)) + B\Big(\frac{1+ir}{2}\Big)\exp(-(1+ir))$

We have the system of equation:

$A\exp(-(1-ir)) + B\exp(-(1+ir)) = 1$

$A \Big(\frac{1-ir}{2}\Big)\exp(-(1-ir))- B \Big(\frac{1+ir}{2}\Big)\exp(-(1+ir)) = 3$

$\Leftrightarrow$

$A\exp(-(1-ir)) + B\exp(-(1+ir)) = 1$

$0 + B\Big(\Big(\frac{1-ir}{2}\Big)\exp(-(1+ir)) - \Big(\frac{1+ir}{2}\Big)\exp(-(1+ir))\Big) = \Big(\frac{1-ir}{2}\Big) - 3$

I understand if you tableflip. I don't know how this can further be simplified. Of course, you can try to verify my calculations for $A$ and $B$ because I haven't done this yet.

Answer 2

Well, assuming the solution to your ODE $$y''-y'+4y=0$$ is of the form $y=e^{ax}$, plug this into the ODE and divide by $e^{ax}$ to get $$a^2-a+4=0,$$ which implies $$a_\pm = \frac{1 \pm \sqrt{1-16}}{2} = \frac{1 \pm i \sqrt{15}}{2}.$$ Hence, $$ y(t) = A \sin\left( t\sqrt{15}/2 \right) + B\cos\left( t\sqrt{15}/2 \right). $$

Can you now use the initial conditions to compute $A$ and $B$?

Answer 3

You can control your answers by inserting in the original equation. For example:

The solution to $y'' - 3y' + 2y = 0$ is $y(x) = c_1e^{x} + c_2e^{2x}$. To control this, one differentiates $y$ and insert in the original equation.

$y' = c_1e^x + 2c_2e^{2x}$,

$y''= c_1e^x + 4c_2e^{2x}$,

$y'' - 3y' + 2y = c_1e^x + 4c_2e^{2x} - 3(c_1e^x + 2c_2e^{2x}) + 2(c_1e^x + c_2e^{2x}) = 0$.

Suppose we have initial values $y(0) = 1$ and $y'(0) = 0$. We can then determine the constants $c_1$ and $c_2$ by

$y(0) = c_1e^{0} + c_2e^{2\cdot 0} = c_1 + c_1 = 1$,

$y'(0) = c_1e^{0} + 2c_2e^{2\cdot 0} = c_1 + 2c_2 = 0 \Leftrightarrow c_1 = -2c_2$.

Solving this, one gets $c_1 = -1$ and $c_2 = 2$.

The above techniques can be applied to control your expression for the general solution and determine $c_1$ and $c_2$.

Good luck.

Answer 4

Hope I don't get the algebra wrong \begin{align} D^{2}-D+4 & = (D-\frac{1}{2})^{2}+4-\frac{1}{4} \\ &=(D-\frac{1}{2})^{2}+\frac{15}{2} \\ &=(D-\frac{1}{2}-i\sqrt{\frac{15}{2}})(D-\frac{1}{2}+i\sqrt{\frac{15}{2}}). \end{align} So the solutions of $(D^{2}-D+4)y=0$ are $$ y = Ae^{-x/2}\sin(\sqrt{15/2}x)+Be^{-x/2}\cos(\sqrt{15/2}x). $$ The conditions are $$ y(-2)=1,\;\; y'(-2)=3. $$ The derivative expression is \begin{align} y' & =A\left(-\frac{1}{2}e^{-x/2}\sin(\sqrt{15/2}x)+e^{-x/2}\sqrt{15/2}\cos(\sqrt{15/2}x)\right) \\ & +B\left(-\frac{1}{2}e^{-x/2}\cos(\sqrt{15/2}x)-e^{-x/2}\sqrt{15/2}\sin(\sqrt{15/2}x)\right) \end{align} Two equations in two unknowns $A$ and $B$ obtained by evaluating $y$ and $y'$ at $x=-2$: $$ \begin{pmatrix} -e\sin(\sqrt{30}) & e\cos(\sqrt{30}) \\ \frac{1}{2}e\sin(\sqrt{30})+e\sqrt{15/2}\cos(\sqrt{30}) & -\frac{1}{2}e\cos(\sqrt{30})+e\sqrt{15/2}\sin(\sqrt{30}) \end{pmatrix}\begin{pmatrix}A \\ B\end{pmatrix}=\begin{pmatrix}1 \\ 3\end{pmatrix} $$ A 2x2 matrix is fairly easy to invert, assuming the determinant is non-zero: $$ \begin{pmatrix}a & b \\ c & d \end{pmatrix}^{-1}= \frac{1}{ad-bc}\begin{pmatrix}d & -b \\ -c & a\end{pmatrix} $$ That gives you $A$ and $B$ with no problem.