User:EGM6321.f11.team1.pan/HW5

= R*5.2 =

Solution
Part (2) Hermite differential equation $$\displaystyle y''-2xy'+2ny=0$$ is not exact if $$\displaystyle n\neq -1$$. And it is in the power form of L2-ODE-VC: $$\displaystyle \alpha x^ry''+\beta x^sy'+\gamma x^ty$$ with $$\displaystyle \alpha=1;\beta=-2;\gamma=2n.$$ $$\displaystyle r=0;s=1;t=0.$$ Assuming that we can find the integrating factor which is in power form, $$\displaystyle h(x,y)=x^my^n$$. Use the exactness condition to find $$m,n\in \mathbb{E}$$, such that the following N2-ODE is exact:
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$$ Rewrite Equation(5,2,1) as $$\displaystyle G(x,y,y',y)=\underbrace{2\alpha x^m y^{n+1}-2x^{m+1}y^ny'}_{g(x,y,p)}+\underbrace{x^my^n}_{f(x,y,p)}y$$ Calculate the following terms in the exactness condition: $$\displaystyle f_x=mx^{m-1}y^n;\,f_{xx}=m(m-1)x^{m-2}y^n$$ $$\displaystyle f_{xy}=mnx^{m-1}y^{n-1};\,f_{xp}=0$$ $$\displaystyle f_y=nx^my^{n-1}$$ $$\displaystyle f_{yy}=n(n-1)x^my^{n-2};\,f_{yp}=0$$ $$\displaystyle g_x=2m\alpha x^{m-1}y^{n+1}-2(m+1)x^my^np;\,g_{xp}=-2(m+1)x^my^n$$ $$\displaystyle g_y=2\alpha (n+1)x^{m}y^{n-1}-2nx^{m+1}y^{n-1}p;\,g_{yp}=-2nx^{m+1}y^{n+1}$$ $$\displaystyle g_p=-2x^{m+1}y^n;\,g_{pp}=0$$ Recall the 2nd exactness condition for L2-ODE-VC:
 * $$\displaystyle x^m y^n(y''-2xy'+2\alpha y)=0$$
 * $$\displaystyle (Equation\;5.2.4)
 * $$\displaystyle (Equation\;5.2.4)
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 * $$\displaystyle f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$$
 * $$\displaystyle (Equation\;5.2.5)
 * $$\displaystyle (Equation\;5.2.5)
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$$ Substituting those calculated terms into Equation(5,2,5) and Equation(5.2.6). The 1st exactness condition Equation(5,2,5) is $$\displaystyle m(m-1)x^{m-2}y^n+2p\cdot mnx^{m-1}y^{n-1}+p^2n(n-1)x^my^{n-2}=-2(m+1)x^my^n+p(-2nx^{m+1}y^{n-1})-2\alpha (n+1)x^my^n+2nx^{m+1}y^{n-1}p$$ The 2nd exactness conditionEquation(5.2.6) is $$\displaystyle 0+p\cdot 0+2nx^my^{n-1}=0$$ It is obvious that only if $$\displaystyle n=0$$, the 2nd exactness condition Equation(5.2.6) is satisfied. Substitute $$\displaystyle n=0$$ into Equation(5,2,5), we can obtain $$\displaystyle m(m-1)x^{m-2}=-2(m+1)x^m-2\alpha x^m$$ It means $$\displaystyle m(m-1)=0$$ and $$\displaystyle m+1+\alpha =0$$ We choose $$\displaystyle m=1,\alpha =-2 \,$$ to make the factor h(x,y) not zero.
 * $$\displaystyle f_{xp}+pf_{yp}+2f_y=g_{pp}$$
 * $$\displaystyle (Equation\;5.2.6)
 * $$\displaystyle (Equation\;5.2.6)
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Part (3) Verify the first three terms of Hermite polynomials. Letting $$\displaystyle y=H_0(x)=1$$ $$\displaystyle y'=H_0'(x)=0$$ $$\displaystyle y=H_0(x)=0$$ Substitute them into the Hermite differential equation Equation(5.2.3): $$\displaystyle 0-2x\cdot 0+0\cdot 1=0$$ Therefore, the first term of Hermite polynomial $$\displaystyle H_0(x)=1$$ is the homogeneous solution of the Hermite differential equation Equation (5.2.3).

Letting $$\displaystyle y=H_1(x)=2x$$ $$\displaystyle y'=H_1'(x)=2$$ $$\displaystyle y=H_1(x)=0$$ Substitute them into the Hermite differential equation Equation(5.2.3): $$\displaystyle 0-2x\cdot 2+2\cdot 2x=0$$ Therefore, the second term of Hermite polynomial $$\displaystyle H_1(x)=2x$$ is the homogeneous solution of the Hermite differential equation Equation (5.2.3).

Letting $$\displaystyle y=H_2(x)=4x^2-2$$ $$\displaystyle y'=H_2'(x)=8x$$ $$\displaystyle y=H_2(x)=8$$ Substitute them into the Hermite differential equation Equation(5.2.3): $$\displaystyle 8-2x\cdot 8x+4\cdot (4x^2-2)=0$$ Therefore, the third term of Hermite polynomial $$\displaystyle H_2(x)=4x^2-2$$ is the homogeneous solution of the Hermite differential equation Equation (5.2.3).

= R*5.4 Find $$y_{xxxxx}$$ in terms of derivatives of y with respect to time= == Given == Transformation of variables: $$\displaystyle x=e^t$$. The derivatives of y with respect to x are: $$\displaystyle y_x=e^{-t} \, y_t$$ $$\displaystyle y_{xx}=e^{-2t} \, (y_{tt}-y_t)$$ $$\displaystyle y_{xxx}=e^{-3t} \, (y_{ttt}-3t_{tt}+2y_t)$$ $$\displaystyle y_{xxxx}=e^{-4t} \,(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)$$

Find
Find $$\displaystyle y_{xxxxx}$$ in terms of the derivatives of y with respect to t.

Solution
Recall that $$\displaystyle \frac{dt}{dx}=\left(\frac{dx}{dt}\right)^{-1}=e^{(-t)}$$ The derivative is
 * $$\displaystyle y_{xxxxx}=\frac{d}{dx}\,(y_{xxxx})=\frac{dy_{xxxx}}{dt}\cdot \frac{dt}{dx}$$
 * $$\displaystyle =\frac{d}{dt}\left[e^{(-4t)}\,(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)\right]\cdot e^{-t}$$
 * $$\displaystyle =e^{-t}\cdot\left[\frac{d}{dt}(e^{-4t}\cdot(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)+e^{-4t}\cdot \frac{d}{dt} (y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)\right]$$
 * $$\displaystyle =e^{-t}\cdot[-4e^{-4t}\cdot (y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)+e^{-4t}\cdot (y_{ttttt}-6y_{tttt}+11y_{ttt}-6y_{tt})$$


 * $$\displaystyle =e^{-5t}(y_{ttttt}-10y_{tttt}+35y_{ttt}-50y_{tt}+24y_t)$$
 * $$\displaystyle$$