- #1

K Murty

- 34

- 17

Hi

As an example of what I am talking about, this method works for this DE:

$$

4y' + 2y = 10\cos(x) \\ \\

10 \cos(x) = \Re( 10 e^{j(x)} ) = \Re(e^{j(x)} \cdot e^{j(0)} ) \rightarrow \text{complex number that captures the amplitude and phase of 10 cos x is} \\ 10 \angle{0} = 10 \\

\text{using the property of exponentials that} {\dfrac{d^{n}}{dx^{n}} e^{j(ax) } = {(aj)}^{n} e^{j(ax) }} \\ \\ \text{Since forcing function is sinusoid y is a sinusoid too of the same angular velocity}

\\

\\ a = 1\\ \text{Angular velocity is one} \\

\\

4J y + 2y = 10\angle{0} \\

\\

y (4J + 2) = 10\angle{0} \\

y = \dfrac{ 10\angle{0} }{2 + 4J } \\ \dfrac{ 10}{ \sqrt{2^2 + 4^2} } \angle{ - \arctan(\frac{4}{2}) } \rightarrow \sqrt{5} \cos(x - \arctan(2) )

\\ \\

y_{p} = \sqrt{5} \cos(x - \arctan(2) )

$$

But it does not work for certain ODE, especially those that I think are in resonance:

$$

y''' - 4y' = 4 \sin(2x)

$$

For the function above, I got the answer as:

$$

y_{p} = \dfrac{1}{4} \cos(2x)

$$

But the wolfram alpha displays the answer for the particular solution as:

$$

\dfrac{ \cos^2(x)}{2}

$$

In this one other example, the method leads to a contradiction:

$$

y'' + y = \cos(x) \\

\text{Solving using method above leads to contradiction} \\ y \cdot 0 = 1

$$

My general questions are, how can I know when to avoid using complex numbers to find the particular solution of ODES?

And secondly, can anyone refer me to something that explains how to solve an ODE in resonance when the forcing function is a trigonometric function (sinusoids only)?

**particular solution only.**As an example of what I am talking about, this method works for this DE:

$$

4y' + 2y = 10\cos(x) \\ \\

10 \cos(x) = \Re( 10 e^{j(x)} ) = \Re(e^{j(x)} \cdot e^{j(0)} ) \rightarrow \text{complex number that captures the amplitude and phase of 10 cos x is} \\ 10 \angle{0} = 10 \\

\text{using the property of exponentials that} {\dfrac{d^{n}}{dx^{n}} e^{j(ax) } = {(aj)}^{n} e^{j(ax) }} \\ \\ \text{Since forcing function is sinusoid y is a sinusoid too of the same angular velocity}

\\

\\ a = 1\\ \text{Angular velocity is one} \\

\\

4J y + 2y = 10\angle{0} \\

\\

y (4J + 2) = 10\angle{0} \\

y = \dfrac{ 10\angle{0} }{2 + 4J } \\ \dfrac{ 10}{ \sqrt{2^2 + 4^2} } \angle{ - \arctan(\frac{4}{2}) } \rightarrow \sqrt{5} \cos(x - \arctan(2) )

\\ \\

y_{p} = \sqrt{5} \cos(x - \arctan(2) )

$$

But it does not work for certain ODE, especially those that I think are in resonance:

$$

y''' - 4y' = 4 \sin(2x)

$$

For the function above, I got the answer as:

$$

y_{p} = \dfrac{1}{4} \cos(2x)

$$

But the wolfram alpha displays the answer for the particular solution as:

$$

\dfrac{ \cos^2(x)}{2}

$$

In this one other example, the method leads to a contradiction:

$$

y'' + y = \cos(x) \\

\text{Solving using method above leads to contradiction} \\ y \cdot 0 = 1

$$

My general questions are, how can I know when to avoid using complex numbers to find the particular solution of ODES?

And secondly, can anyone refer me to something that explains how to solve an ODE in resonance when the forcing function is a trigonometric function (sinusoids only)?

Last edited: