User:Egm4313.s12.team13.rlh.r2

Problem Statement
5. Find a general solution. Check by substitution.

$$  \; {y}''+2\pi {y}'+\pi ^{2}y=0 $$

Solution
$$  \; {y}''+2\pi {y}'+\pi ^{2}y= \frac{d^2y}{dx^2}\ + 2 \pi \frac{dy}{dt}\ + \pi ^{2}y=0   $$ (4.0a)

If $$ \; \frac{d}{dx}\ = \lambda $$,

The homogeneous linear ODE with constant coefficients becomes:

$$ \; \lambda ^{2}+2\pi \lambda +\pi ^{2}=0 $$ (4.1a)

By using the quadratic formula, or in this case factoring, we find the solutions of the equation:

Solving for $$ \; \lambda $$,

$$ \; (\lambda +\pi) (\lambda +\pi)=0 $$ (4.2a)

$$ \; \lambda =-\pi, -\pi $$

General Solution:

General solution in the case of distinct real roots:

$$  \; y=c_1e ^ { \lambda _1 x} + c_2e ^ { \lambda _2 x}  $$ (4.3a)

So the general solution becomes:

$$  \; y=c_1e ^ {-\pi x} + c_2xe ^ {-\pi x}= e ^ {-\pi x} (c_1+c_2x) $$ (4.4a)

Check by Substitution:

To check by substitution, you need to find $$  \; y, y', y'' $$ and plug them into the original ODE.

$$  \; y= e ^ {-\pi x} (c_1+c_2x) $$

$$  \; y'= -\pi c_1e ^ {-\pi x} + c_2 (e ^ {-\pi x} -\pi xe ^ {-\pi x})= - \pi e ^{- \pi x} (c_1 + c_2x) + c_2 e^ {- \pi x} $$(4.5a)

$$  \; y''= \pi ^{2} c_1e ^ {- \pi x} - \pi c_2e ^ {-\pi x} -\pi c_2(e ^ {- \pi x} - \pi xe^ {- \pi x} )= \pi ^ {2} e^ {- \pi x} (c_1+ c_2x) -2 \pi c_2 e^ {- \pi x} $$ (4.6a)

Plugging in  $$   \; y, y', y $$ into $$   \; {y}+2\pi {y}'+\pi ^{2}y=0 $$ :

$$  \; [ \pi ^ {2} e^ {- \pi x} (c_1 + c_2x) - 2 \pi c_2e^ {- \pi x}] + 2 \pi [ - \pi e^ {- \pi x} (c_1 + c_2x) + c_2e^ {- \pi x} ] + \pi ^ {2} [ e^ {- \pi x} (c_1+c_2x)]=0 $$

When plugging in, the general solution is confirmed because it equals zero.

Problem Statement
6. Find the general solution. Check by substitution.

$$ 10y''-32y'+25.6y=0 $$

Solution
The original problem can be re-written as:
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$$  \displaystyle 10\frac{d^2y}{dx^2}-32\frac{dy}{dx}+25.6y=0 $$     (4.0b)
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then let
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$$  \displaystyle \frac{d}{dx}=\lambda $$     (4.1b) By substituting equation 4.1b into 4.0b we get:
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$$  \displaystyle 10\lambda^2-32\lambda+25.6=0 $$     (4.2b) Solve for $$ \; \lambda $$ using the quadratic equation.
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Quadratic equation:
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$$  \displaystyle \lambda=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$     (4.3b) By using equation 4.3b it is shown that $$ \lambda=\frac{8}{5}$$
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Since $$ \; \lambda$$ is a solution to the ODE then
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$$  \displaystyle y=c_1e ^ { \lambda _1 x} + c_2e ^ { \lambda _2 x} $$ (4.4b) is the solution of the ODE with real and distinct roots.
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By plugging the value of $$ \; \lambda $$ into equation 4.4b we come to the general solution.

General Solution $$ y=e^{\frac{8}{5}x}({c_1+c_2 x}) $$

Checking by Substitution To check by substitution, you need to find $$ \; y, y', y''$$ and plug them into the original ODE.


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$$  \displaystyle y'=\frac{8}{5}e^{\frac{8}{5}x}(c_1+c_2 x)+c_2 e^{\frac{8}{5}x} $$     (4.5b)
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$$  \displaystyle y''=\frac{64}{25}e^{\frac{8}{5}x}(c_1+c_2 x)+\frac{16}{5}c_2 e^{\frac{8}{5}x} $$     (4.6b) Since we already know what $$ \; y $$ is equal to from our general solution we are ready to substitute the general solution, equation 4.5b, and 4.6b back into the original equation.
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After substitution we get:
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$$  \displaystyle \frac{64}{25}e^{\frac{8}{5}x}(c_1+c_2 x)-\frac{64}{25}e^{\frac{8}{5}x}(c_1+c_2 x)+32c_2e^{\frac{8}{5}x}-32c_2e^{\frac{8}{5}x}=?? $$ This equation clearly equals  zero  which proves our general solution is  correct 
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Problem Statement
8. Find the general solution. Check by substitution.

$$ y''+y'+3.25y=0 $$

Solution
The original problem can be re-written as:
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$$  \displaystyle \frac{d^2y}{dx^2}+\frac{dy}{dx}+3.25y=0 $$     (8.0a)
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then let
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$$  \displaystyle \frac{d}{dx}=\lambda $$     (8.1a) By substituting equation 8.1a into 8.0a we get:
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$$  \displaystyle \lambda^2+\lambda+3.25=0 $$     (8.2a) Solve for $$ \; \lambda $$ using the quadratic equation.
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Quadratic equation:
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$$  \displaystyle \lambda=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$     (8.3a) By using equation 8.3a it is shown that $$ \lambda=\frac{-1\pm i \sqrt{12}}{2}$$
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Since $$ \; \lambda$$ is a solution to the ODE then
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$$  \displaystyle y=e^{-\frac{1}{2}x}(c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x) $$     (8.4a) is the solution of the ODE with real and distinct roots.
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General Solution $$ y=e^{-\frac{1}{2}x}(c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x) $$

Checking by Substitution To check by substitution, you need to find $$ \; y, y', y''$$ and plug them into the original ODE.


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$$  \displaystyle y'=-\frac{1}{2}e^{-\frac{1}{2}x}(c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x)+e^{-\frac{1}{2}x}(-\sqrt{12}c_1 \sin\sqrt{12}x+\sqrt{12}c_2 \cos\sqrt{12}x) $$     (8.5a)
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$$  \displaystyle y''=\frac{1}{4}e^{-\frac{1}{2}x}(c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x)-\frac{1}{2}e^{-\frac{1}{2}x}(-\sqrt{12}c_1 \sin\sqrt{12}x+\sqrt{12}c_2 \cos\sqrt{12}x)-\frac{1}{2}e^{-\frac{1}{2}x}(-\sqrt{12}c_1 \sin\sqrt{12}x+\sqrt{12}c_2 \cos\sqrt{12}x)$$
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$$+\frac{1}{2}e^{-\frac{1}{2}x}(-12c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x) $$     (8.6a) Since we already know what $$ \; y $$ is equal to from our general solution we are ready to substitute the general solution, equation 8.5a, and 8.6a back into the original equation.
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After substitution we get:
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$$  \displaystyle y''+y'+3.25y=0 $$ So the solution to the ODE is: $$ y=e^{-\frac{1}{2}x}(c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x) $$
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