User:Egm6321.f12.team4.harris/Homework5 R*6.2

Problem R*6.2 – Solving an Euler (homogeneous) L2-ODE-VC
sec31-3

Statement
Solve the Euler (homogeneous) linear second order ordinary differential equation with varying coefficients (L2-ODE-VC)
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$$ x^2y''-2xy'+2y=0 $$     (6.2.1)
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using Method 2 (sec31-4 ) with the trial solution $$ y=x^r $$, such that $$ r $$ is a constant, with the boundary conditions $$ y(1) = 3 y(2) = 4 $$. Plot the results over the domain of the L2-ODE-VC.

Solution
In Method 2 (see link above), one must begin with $$ y=x^r $$ and take the proper derivatives.


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$$ y'=rx^{r-1} $$     (6.2.2)
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$$ y''=r(r-1)x^{r-2} $$     (6.2.3)
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Once the derivatives are computed, they must then be substituted back into equation 6.2.1. This allows one to determine a characteristic equation and solve for the roots of the equation.


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$$ (x^2)r(r-1)x^{r-2}-(2x)rx^{r-1}+2x^r=0 $$     (6.2.4)
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Adding exponents of like bases, one can reduce equation 6.2.4 to


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$$ r(r-1)x^r-2rx^r+2x^r=0 $$     (6.2.5)
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Dividing out the $$ x^r $$ from each term, one can see that the characteristic equation becomes


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$$ r(r-1)-2r+2=0 $$     (6.2.6)
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Equation 6.2.6 simplifies to


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$$ r^2-3r+2=0 $$     (6.2.7)
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Equation 6.2.7 can then be solved to get the two roots, $$ r_1=1 $$ and $$ r_2=2 $$. These roots are then substituted into the general solution


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$$ y(x)=c_1x^{r_1}+c_2x^{r_2} $$     (6.2.8)
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where $$ c_1 $$ and $$ c_2 $$ are constants that can be solved for via boundary conditions. Plugging the roots into equation 6.2.8 and using the boundary conditions in the problem statement, one can see that


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$$ y(x)=c_1x^1+c_2x^2 $$     (6.2.9)
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$$ y(1)=c_1(1)+c_2(1)=3 $$     (6.2.10)
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$$ y(2)=c_1(2)+c_2(4)=4 $$     (6.2.11)
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Multiplying equation 6.2.10 by 2 and subtracting equation 6.2.11 from it, one can see that $$ c_1 = 4 $$ and $$ c_2 = -1 $$. Thus


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$$ y(x)=4x-x^2 $$     (6.2.12)
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Equation 6.2.12 is plotted over the domain ?????. This plot is shown below.

Author and References

 * Solved and typed by -- Kaitlin Harris