User:EGM4313.s12.team4.jones/R2.4

= Problem 4a: General Solution for ODEs =

Problem Statement
Problem found in the textbook on page 59, Problem 5 Find a general solution. Check your answer by substitution.


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$$\displaystyle {y}''+2\pi {y}'+\pi^{2}y=0$$
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Solution
Using the following:


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$$\displaystyle {y}=e^{rx}$$
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$$\displaystyle {y}'=re^{rx}$$
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$$\displaystyle {y}''=r^2e^{rx}$$
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Now plug in $$\displaystyle {y},{y}',{y}''$$ into the original equation:
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$$\displaystyle r^2e^{rx}+2\pi{r}e^{rx}+\pi^{2}e^{rx}=0$$
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Now factor out $$\displaystyle e^{rx}$$:
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$$\displaystyle e^{rx}(r^2+2{\pi}r+{\pi}^2)=0$$
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Since $$\displaystyle e^{rx}=0$$, the ODE can be expressed as follows:
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$$\displaystyle r^2+2{\pi}r+{\pi}^2=0$$
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Solving
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$$\displaystyle (r+{\pi})^2=0$$
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$$\displaystyle r=-\pi,-\pi$$
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The final general solution for this problem is:


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Check by substitution:


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$$\displaystyle {y}'=-\pi c_{1}e^{-\pi x}+c_{2}[e^{-\pi x}-\pi xe^{-\pi x}]$$
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$$\displaystyle {y}''= {\pi}^2c_{1}e^{-\pi x}+c_{2}[-\pi e^{-\pi x}-\pi e^{-\pi x}+{\pi}^2xe^{-\pi x}]$$
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Substituting $$\displaystyle y, {y}', {y}''$$, we reach:
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$$\displaystyle \pi^{2}c_{1}e^{-\pi x}+c_{2}[-2\pi e^{-\pi x}+{\pi}^2xe^{-\pi x}]+2\pi[-\pi c_{1}e^{-\pi x}+c_{2}(e^{-\pi x}-\pi xe^{-\pi x})]+{\pi}^2[c_{1}e^{-\pi x}+c_{2}xe^{-\pi x}] =0$$
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$$\displaystyle \pi^{2}c_{1}e^{-\pi x}+\pi^{2}c_{2}xe^{-\pi x} -2\pi c_{2}e^{-\pi x}-2\pi^{2}c_{1}e^{-\pi x} -2\pi^{2} c_{2}xe^{-\pi x}+2\pi c_{2}e^{-\pi x} +\pi^{2}c_{1}e^{-\pi x} +\pi^{2}c_{2}xe^{-\pi x}=0$$
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All terms cancel on the left side of the equation giving:


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$$\displaystyle 0=0$$
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Therefore the solution is correct.

= Problem 4b: General Solution for ODEs =

Problem Statement
Problem found in the textbook on page 59, Problem 6 Find a general solution. Check your answer by substitution.


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$$\displaystyle 10{y}''-32{y}'+25.6y=0$$
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Solution
Using the following:


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$$\displaystyle {y}=e^{rx}$$
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$$\displaystyle {y}'=re^{rx}$$
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$$\displaystyle {y}''=r^2e^{rx}$$
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Now plug in $$\displaystyle {y},{y}',{y}''$$ into the original equation:
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$$\displaystyle 10r^2e^{rx}-32{r}e^{rx}+25.6e^{rx}=0$$
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Now factor out $$\displaystyle e^{rx}$$:
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$$\displaystyle e^{rx}(10r^2-32r+25.6)=0$$
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Since $$\displaystyle e^{rx}=0$$, the ODE can be expressed as follows:
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$$\displaystyle 10r^2-32r+25.6=0$$
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Solving
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$$\displaystyle r=1.6$$
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The final general solution for this problem is:


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Check by substitution:


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$$\displaystyle {y}'=1.6c_{1}e^{1.6x}+c_{2}[e^{1.6x}-1.6xe^{1.6x}]$$
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$$\displaystyle {y}''= {1.6}^2c_{1}e^{1.6x}+c_{2}[1.6e^{1.6x}-1.6e^{1.6x}+{1.6}^2xe^{1.6x}]$$
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Substituting $$\displaystyle y, {y}', {y}''$$, we reach:
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$$\displaystyle 25.6c_{1}e^{1.6x}+10c_{2}[3.2e^{1.6x}+2.56xe^{1.6x}]-51.2c_{1}e^{1.6x}-32c_{2}(e^{1.6x}+1.6xe^{1.6x})]+25.6[c_{1}e^{1.6x}+c_{2}xe^{1.6x}] =0$$
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$$\displaystyle 25.6c_{1}e^{1.6x}+32c_{2}e^{1.6x}+25.6c_{2}xe^{1.6x}-51.2c_{1}e^{1.6x}-32c_{2}e^{1.6x}-51.2c_{2}xe^{1.6x}+25.6c_{1}e^{1.6x}+25.6c_{2}xe^{1.6x} =0$$
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All terms cancel on the left side of the equation giving:


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$$\displaystyle 0=0$$
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Therefore the solution is correct.