User:Egm4313.s12.team6.jagolinzer.r2

My problems for Report 2: 2.7 and 2.8

Given:
$$e^t$$ $$\cos t$$ $$\sin t$$

Find:
Develop the MacLaurin Series which is the Taylor Series at $$t=0$$  for the given equations.

Solution:
The MacLaurin series is of the form: $$f(x)=f(0)+f'(0)+\frac{f''(0)}{2!}+\frac{f^{(3)}(0)}{3!}+...$$

For  $$e^t$$   :
$$f(x)=e^0+xe^0+\frac{x^2e^0}{2!}+\frac{x^3e^0}{3!}+...$$

This simplifies to:

$$f(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$

And a Summation for this series can be derived as:

$$\sum_{i=0}^{n}\frac{x^i}{i!}$$

For  $$\cos t$$   :
$$f(x)=\cos 0-x\sin0+\frac{x^2\cos 0}{2!}+\frac{x^3\sin 0}{3!}+...$$

This simplifies to:

$$f(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$$

And a Summation for this series can be derived as:

$$\sum_{i=0}^{n}\frac{x^{2i}}{(2i)!}$$

For  $$\sin t$$   :
$$f(x)=\sin 0-x\cos0+\frac{x^2\sin 0}{2!}+\frac{x^3\cos 0}{3!}+...$$

This simplifies to:

$$f(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$$

And a Summation for this series can be derived as:

$$\sum_{i=0}^{n}\frac{x^{2i+1}}{(2i+1)!}$$

R2.8
This problem consist of problems 8 and 15 from the tenth edition of Erwin Kreyszig's Advanced Engineering Mathematics in section 2.2 on page 59.

Given:
The ODE:    $$y''+y'+3.25y=0$$

Find:
The general solution. Check by substitution.

Solution:
Let:    $$y=e^{\lambda x}$$

Taking the first derivative we get:    $$y'=\lambda e^{\lambda x}$$

Taking the second derivative we get:    $$y''=\lambda^2 e^{\lambda x}$$

Plugging this into   $$y''+y'+3.25y=0$$    we get:     $$e^{\lambda x}(\lambda^2+\lambda+3.25)=0$$

Since  $$e^{\lambda x}\neq 0  \therefore (\lambda^2+\lambda+3.25)=0$$

Using the Quadratic formula $$\frac{1}{2}(-a\pm \sqrt{a^2-4ab})$$ we get  $$a^2-4b<0$$   and therefore it is complex conjugate roots.

Our roots are therefore  $$y_{1}=e^{\frac{-x}{2}}(\cos \omega x)$$   and   $$y_{1}=e^{\frac{-x}{2}}(\sin \omega x)$$

Using $$\omega^2=b-\frac{a^2}{4}=3.25-0.25=3$$ we get  $$\omega=\sqrt{3}$$

Therefore the general solution is:


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$$y=e^{\frac{-x}{2}}(A\cos \sqrt{3}x+B\sin \sqrt{3}x)$$ Now we must check our solution by taking the derivatives of the general solution and plugging them into the ODE.
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$$y'=e^{\frac{-x}{2}}(-\omega A\sin \omega x+\omega B\cos \omega x-\frac{A}{2}\cos \omega x-\frac{B}{2}\sin \omega x)$$

$$y''=e^{\frac{-x}{2}}(\frac{\omega A}{2}\sin \omega x-\frac{\omega B}{2}\cos \omega x+\frac{A}{4}\cos \omega x+\frac{B}{4}\sin \omega x-\omega^2 A\cos \omega x-\omega^2 B\sin \omega x+\frac{\omega A}{2}\sin \omega x-\frac{\omega B}{2}\cos \omega x)$$

plugging into  $$y''+y'+3.25y=0$$   with the above we get:

$$0=e^{\frac{-x}{2}}(0)$$  which confirms our solution.

Given:
The ODE:    $$y''+0.54y'+(0.0729+\pi)y=0$$

Find:
The general solution. Check by substitution.

Solution:
Let:    $$y=e^{\lambda x}$$

Taking the first derivative we get:    $$y'=\lambda e^{\lambda x}$$

Taking the second derivative we get:    $$y''=\lambda^2 e^{\lambda x}$$

Plugging this into   $$y''+0.54y'+(0.0729+\pi)y=0$$    we get:     $$e^{\lambda x}(\lambda^2+0.54\lambda+(0.0729+\pi))=0$$

Since  $$e^{\lambda x}\neq 0  \therefore (\lambda^2+0.54\lambda+(0.0729+\pi))=0$$

Using the Quadratic formula $$\frac{1}{2}(-a\pm \sqrt{a^2-4ab})$$ we get  $$a^2-4b<0$$   and therefore it is complex conjugate roots.

Our roots are therefore  $$y_{1}=e^{-0.27x}(\cos \omega x)$$   and   $$y_{1}=e^{-0.27x}(\sin \omega x)$$

Using $$\omega^2=b-\frac{a^2}{4}=(0.0729+\pi)-\frac{0.54^2}{4}=\pi$$ we get  $$\omega=\sqrt{\pi}$$

Therefore the general solution is:


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$$y=e^{-0.27x}(A\cos \sqrt{\pi}x+B\sin \sqrt{\pi}x)$$ Now we must check our solution by taking the derivatives of the general solution and plugging them into the ODE.
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$$y'=e^{-0.27x}(-\omega A\sin \omega x+\omega B\cos \omega x-0.27A\cos \omega x-0.27B\sin \omega x)$$

$$y''=e^{-0.27x}(0.27\omega A\sin \omega x-0.27\omega B\cos \omega x+0.0729\omega A\cos \omega x+0.0729\omega A\sin \omega x-\omega^2 A\cos \omega x-\omega^2 B\sin \omega x+0.27\omega A\sin \omega x-0.27\omega B\cos \omega x)$$

plugging into  $$y''+0.54y'+(0.0729+\pi)y=0$$   with the above we get:

$$0=e^{-0.27x}(0)$$  which confirms our solution.