User:Egm4313.f13.team9.shankwitz

Problem Statement
Solve the initial value problem and graph the solution over the intervals $$\displaystyle 0\leq x \leq 1\,, \, 0\leq x\leq 5 $$
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$$  y''+9y'+20=0 $$     (1)
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Given
$$ \displaystyle y(0)=1 $$ $$ \displaystyle y'(0)=\frac{1}{2} $$

Step 1: Find a General Solution
The ODE is a linear, second-order, homogeneous differential equation with constant coefficients. So, the following equation was chosen as a solution.
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$$ y=e^{\lambda x} $$ (2) The first and second derivatives are as follows:
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$$ y'=\lambda e^{\lambda x} $$ (3)
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$$ y''=\lambda ^2 e^{\lambda x} $$ (4) Plugging the solution and its derivatives back into the original ODE, we receive
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$$ (\lambda ^2 +9\lambda +20)e^{\lambda x}=0 $$ (5) and the characteristic equation
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$$ \lambda ^2 +9\lambda +20=0 $$ (6)
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This gives us 2 real solutions from the quadratic formula, $$ \displaystyle \lambda_1 =-4 \,, \, \lambda_2 =-5 $$ and the general solution:
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$$ y=c_1 e^{-4x}+c_2e^{-5x} $$ (7)
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Step 2: Solve the IVP
Equation (7) and its derivative
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$$ y'=-4c_1 e^{-4x}-5c_2e^{-5x} $$ (8)
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can be set equal to the initial values given


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$$ y(0)=1=c_1 +c_2 $$ (9)
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$$ y'(0)=\frac{1}{2}=-4c_1-5c_2 $$ (10)
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Solving (9) and (10) simultaneously gives us the c-values and the solution to the IVP
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$$ y=5.5e^{-4x}-4.5e^{-5x} $$ (11)
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Step 3: Check Answer with Substitution
Our solution and its first two derivatives can be substituted into the original ODE
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$$ y''+8y'+20y=0 $$ (12)
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$$ y=5.5e^{-4x}-4.5e^{-5x} $$ (13)
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$$ y'=-22e^{-4x}+22.5e^{-5x} $$ (14)
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$$ y''=88e^{-4x}-112.5e^{-5x} $$ (15)
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$$ [88+9(-22)+20(5.5)]e^{-4x}+[-112.5+9(22.5)+20(-4.5)]e^{-5x}=0 $$ (16) Which is true.
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