User:EGM6321.f12.team4.Rui.C/Homework6 R6.4(1.3)(1.4)

Statement
Consider the characteristic equation:

where $$\lambda=5$$ ,

(1) Euler L2-ODE-VC:


 * (1.1) Find $$a_2,a_1,a_0$$ such that (6.4.0.1) is characteristic equation of (6.4.0.2).


 * (1.2) 1st homogeneous solution: $$y_1(x)=x^\lambda$$


 * (1.3) Complete solution: find c(x) such that


 * (1.4) Find the 2nd homogeneous solution $$y_2(x)$$

(2) L2-ODE-CC

Repeat the steps.

Solution(1.1)
Given,


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$$ a_2x^2y''+a_1 xy'+a_1y=0 $$     (6.4.1.1)
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We start by transforming the variable x to t, by choosing $$ x=e^t $$


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$$ \frac{dy}{dx}= (\frac{d}{dt} \frac{dt}{dx})y $$     (6.4.1.2)
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Since $$\frac{dt}{dx} = e^{-t} $$


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$$ \frac{dy}{dx}= (e^{-t} \frac{d}{dt})y $$     (6.4.1.3)
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To calculate $$ y'' $$,


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$$ \frac{d}{dx}(\frac{dy}{dx})= (\frac{d}{dt} \frac{dt}{dx})(\frac{dy}{dx}) $$     (6.4.1.4)
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$$ \frac{d}{dx}(\frac{dy}{dx})= (e^{-t} \frac{d}{dt})(e^{-t} \frac{d}{dt})y $$     (6.4.1.5)
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Differentiating by parts,
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$$ \frac{d}{dx}(\frac{dy}{dx})= (e^{-2t} \frac{dy}{dt}) + (e^{-2t} \frac{d^2y}{dt^2}) $$     (6.4.1.6)
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Substituting (6.4.1.4) & (6.4.1.6) into (6.4.1.1),


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$$ x^2 e^{-2t}a_2(y_{tt}-y_t) + a_1xe^{-t}y_t + a_0y =0 $$     (6.4.1.7)
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Substituting $$ x=e^t $$ & $$ y=e^{rt} $$ the characteristic equation becomes,


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$$ a_2r^2 + (-a_2+a_1)r + a_0 =0 $$     (6.4.1.8)
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For the characteristic equation to be $$ (\lambda -5)^2=0$$, the following equations should be same,


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$$ a_2r^2 + (-a_2+a_1)r + a_0 =0 $$
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$$ r^2 - 10r +25 =0 $$
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Comparing the above,


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$$ a_2 = 1 $$
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$$a_1 = -9 $$
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$$a_0= 25 $$
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Solution(1.2)
The Euler L2-ODE VC becomes,

Substituting $$ y = x^\lambda $$ ,

which gives, $$ (\lambda -5)^2=0 $$,

Hence, the first homogeneous solution is,

Solution(1.3)
The variation of parameters:

Substitute it into the Euler L2-ODE-VC:

We have:

It can be further reorganized as:

where $$k_1, k_2$$ are constants.

Solution(1.4)
Substitute (6.4.3.7) and expression for $$y_1(x)$$ into (6.4.0.3), we have:

Then it is obvious that the second term is the first homogeneous solution and the first term is the second homogeneous solution.

Solution(2.1)
Now,we consider the Euler L2-ODE CC,

Start by choosing $$ y= e^rx $$,

Substituting (6.4.2.2) and (6.4.2.3) into (6.4.2.1)

which can be rewritten as,

Comparing with the characteristic equation,

we get ,

Solution(2.2)
The first homogeneous solutions can be found from as we know the value of r from the characteristic equation,

Solution(2.3)
From the variation of parameters,

Substituting in Eq. (6.4.2.1)

which gives

From this we get,

Hence the complete solution is ,

Solution(2.4)
From (6.4.2.9) we can identify the second term $$e^{5x} $$ as the first homogeneous solution. The second homogeneous solution is thus $$ x e^{5x} $$

Author and References

 * Solved and Typed by -- Pushkar Mishra and Rui Che