User:Egm4313.s12.team4.blj/editpage

= Problem 3: General Solution for ODEs =

Problem Statement
Problem found in the textbook on page 59, problems 3 and 4. Find a general solution. Check your answer by substitution.


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a) $$\displaystyle {y}''+6 {y}'+8.96y=0$$
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 * (2.3a.1)
 * }


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b) $$\displaystyle {y}''+4{y}'+(pi^{2}+4)y=0$$
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 * (2.3b.1)
 * }

Solution (2.3a)
Using the following:


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$$\displaystyle {y}'=\frac{dy}{dx}$$
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 * (2.3a.2)
 * }


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$$\displaystyle {y}''=\frac{d^{2}y}{dx^{2}}$$
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 * (2.3a.3)
 * }

The given equation can be rewritten as follows, using (2.3a.2) and (2.3a.3):
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$$\displaystyle \frac{d^{2}y}{dx^{2}}+6\frac{dy}{dx}+8.96y=0$$
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 * (2.3a.4)
 * }

Substituting
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$$\displaystyle D= \frac{d}{dx}$$
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 * (2.3a.5)
 * }

The ODE can be expressed as follows:
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$$\displaystyle (D^{2}+6D+8.96)y=0$$
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 * (2.3a.6)
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Solving
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$$\displaystyle (D^{2}+6D+8.96)=0$$
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 * (2.3a.7)
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Use the quadratic formula


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$$\displaystyle \frac{6\pm\sqrt{6^{2}-4(1)(8.96)}}{2(1)} $$
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 * (2.3a.8)
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Arriving at


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$$\displaystyle (D+3.2)(D+2.8)=0$$
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 * (2.3a.9)
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$$\displaystyle D=-3.2,-2.8$$
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 * (2.3a.10)
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The final general solution for this problem is:

Reach the general solution :


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 * (2.3a.11)
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Check by substitution:

From (2.3a.11), we find,
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$$\displaystyle {y}'=-3.2C_{1}e^{-3.2x}-2.8C_{2}e^{-2.8x}$$
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 * (2.3a.12)
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$$\displaystyle {y}''=10.4C_{1}e^{-3.2x}+7.84C_{2}e^{-2.8x}$$
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 * (2.3a.13)
 * }

Substituting (2.3a.11), (2.3a.12), and (2.3a.13) in (2.3a.1), we reach:
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$$\displaystyle (10.4C_{1}e^{-3.2x}+7.84C_{2}e^{-2.8x})+6(10.4C_{1}e^{-3.2x}+7.84C_{2}e^{-2.8x})+8.96(C_{1}e^{-3.2x}+C_{2}e^{-2.8x})=0$$
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 * (2.3a.14)
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$$\displaystyle 19.2C_{1}e^{-3.2x}+16.8C_{2}e^{-2.8x} -19.2C_{1}e^{-3.2x}-16.8C_{2}e^{-2.8x}=0$$
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 * (2.3a.15)
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$$\displaystyle 0=0$$
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 * <p style="text-align:right">(2.3a.16)
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Therefore the solution is correct.

Solution (2.3b)
Using the following:


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$$\displaystyle {y}'=\frac{dy}{dx}$$
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 * style="width:95%" |
 * <p style="text-align:right">(2.3b.2)
 * }


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$$\displaystyle {y}''=\frac{d^{2}y}{dx^{2}}$$
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 * style="width:95%" |
 * <p style="text-align:right">(2.3b.3)
 * }

The given equation can be rewritten as follows, using (2.3b.2) and (2.3b.3):
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$$\displaystyle \frac{d^{2}y}{dx^{2}}+4\frac{dy}{dx}+(\pi^2 +4)y=0$$
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 * style="width:95%" |
 * <p style="text-align:right">(2.3b.4)
 * }

Substituting
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$$\displaystyle D= \frac{d}{dx}$$
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 * <p style="text-align:right">(2.3b.5)
 * }

The ODE can be expressed as follows:
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$$\displaystyle (D^{2}+4D+(\pi^2 +4))y=0$$
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 * <p style="text-align:right">(2.3b.6)
 * }

Solving
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$$\displaystyle (D^{2}+4D+(\pi^2 +4))=0$$
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 * <p style="text-align:right">(2.3b.7)
 * }

Use the quadratic formula


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$$\displaystyle \frac{4\pm\sqrt{4^{2}-4(1)(\pi^2 +4)}}{2(1)} $$
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 * <p style="text-align:right">(2.3b.8)
 * }

Arriving at


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$$\displaystyle (D+(2+\pi i))(D+(2-\pi i)=0$$
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 * <p style="text-align:right">(2.3b.9)
 * }


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$$\displaystyle D=-(2+\pi i),-(2-\pi i)$$
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 * <p style="text-align:right">(2.3b.10)
 * }

The final general solution for this problem is:

Reach the general solution :


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 * <p style="text-align:right">(2.3b.11)
 * }

Check by substitution:

From (2.3b.11), we find,
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$$\displaystyle {y}'= -2(A \cos (\pi x) + B \sin (\pi x))e^{-2x} + (-A \pi \sin (\pi x) + B \pi \cos (\pi x))e^{-2x}$$
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 * <p style="text-align:right">(2.3b.12)
 * }


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$$\displaystyle {y}''=4(A \cos (\pi x) + B \sin (\pi x))e^{-2x} - 2(-A \pi \sin (\pi x) + B \pi \cos (\pi x))e^{-2x}$$ $$\color{White}y''= \color{Black} - 2(-A \pi \sin (\pi x) + B \pi \cos (\pi x))e^{-2x} + (-A \pi ^2 \cos (\pi x) - B \pi ^2 \sin (\pi x))e^{-2x}$$
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 * <p style="text-align:right">(2.3b.13)
 * }

Substituting (2.3b.11), (2.3b.12), and (2.3b.13) in (2.3b.1), we reach:
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$$\displaystyle 4(A \cos (\pi x) + B \sin (\pi x))e^{-2x} - 2(-A \pi \sin (\pi x) + B \pi \cos (\pi x))e^{-2x} - 2(-A \pi \sin (\pi x) + B \pi \cos (\pi x))e^{-2x} $$ $$\displaystyle+ (-A \pi ^2 \cos (\pi x) - B \pi ^2 \sin (\pi x))e^{-2x} +4 -2(A \cos (\pi x) + B \sin (\pi x))e^{-2x}) + (-A \pi \sin (\pi x) + B \pi \cos (\pi x))e^{-2x}$$ $$\displaystyle+(\pi^{2}+4)((A\cos(\pi x) + B\sin(\pi x))e^{-2x})=0$$
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 * <p style="text-align:right">(2.3b.14)
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$$\displaystyle 0=0$$
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 * <p style="text-align:right">(2.3b.15)
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Therefore the solution is correct.