User:EGM6321.f11.team1.pan/HW6

= R*6.1 Special IFM to solve a nonhomogeneous L2-ODE-CC with general f(t)= == Given == A nonhomogeneous L2-ODE-CC is
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle a_2y''+a_1y'+a_0y=f(t)$$
 * $$\displaystyle (Equation\;6.1.1)
 * $$\displaystyle (Equation\;6.1.1)
 * }
 * }

Find
$$\mathbf{(1)}$$ Find the PDEs that govern the integrating factor h(x,y) for Equation(6.1.1).

$$\mathbf{(2)}$$ Trial solution for the integrating factor $$\displaystyle h(t)=e^{\alpha t}$$ where $$\displaystyle \alpha$$ is unknown to be determined.
 * {| style="width:100%" border="0" align="left"

$$ Because of the integrating factor in exponential form, assume the LHS of Equation(6.1.2) take the form :
 * $$\displaystyle \int \, e^{\alpha t}[a_2y''+a_1y'+a_0y]dt= \int \, e^{\alpha t}f(t)dt$$
 * $$\displaystyle (Equation\;6.1.2)
 * $$\displaystyle (Equation\;6.1.2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle \int \, e^{\alpha t}[a_2y''+a_1y'+a_0y]dt= e^{\alpha t}[\bar a_1y'+\bar a_0y]$$
 * $$\displaystyle (Equation\;6.1.3)
 * $$\displaystyle (Equation\;6.1.3)
 * }
 * }

$$\mathbf{(2.1)}$$ Find $$\displaystyle (\bar a_1,\bar a_0)$$ in terms of $$\displaystyle (a_0,a_1,a_2)$$.

$$\mathbf{(2.2)}$$ Find the quadratic equation for $$\displaystyle \alpha$$
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle a_2\alpha ^2-a_1\alpha+a_0=0$$
 * $$\displaystyle (Equation\;6.1.4)
 * $$\displaystyle (Equation\;6.1.4)
 * }
 * }

$$\mathbf{(2.3)}$$ Reduced-order equation: Equation(6.1.3)and Equation(6.1.4) lead to
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle \bar a_1y'+\bar a_0y=e^{-\alpha t}\int d^{\alpha t}f(t)dt$$
 * $$\displaystyle (Equation\;6.1.5)
 * $$\displaystyle (Equation\;6.1.5)
 * }
 * }

$$\mathbf{(2.4)}$$ Use the IFM to solve Equation(6.1.5)
 * {| style="width:100%" border="0" align="left"

$$ Find the solution y(t) for general excitation f(t).
 * $$\displaystyle \bar h(t)=exp[\int \, \frac{\bar a_0}{\bar a_1}dt]=e^{\beta t}$$
 * $$\displaystyle (Equation\;6.1.6)
 * $$\displaystyle (Equation\;6.1.6)
 * }
 * }

$$\mathbf{(2.5)}$$ Show that
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle \alpha\beta=\frac{a_0}{a_2}$$
 * $$\displaystyle (Equation\;6.1.7)
 * $$\displaystyle (Equation\;6.1.7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ thus $$\displaystyle (\alpha,\beta)$$ are roots of the quadratic equation:
 * $$\displaystyle \alpha+\beta=\frac{a_1}{a_2}$$
 * $$\displaystyle (Equation\;6.1.8)
 * $$\displaystyle (Equation\;6.1.8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle (\lambda-\alpha)(\lambda-\beta)=\lambda^2-\underbrace{(\alpha+\beta)}_{\displaystyle \frac{a_1}{a_2}}\lambda+\underbrace{\alpha\beta}_{\displaystyle \frac{a_0}{a_2}}=0$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.9)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.9)
 * }
 * }

$$\mathbf{(2.6)}$$ Deduce the particular solution $$y_p(t)\,$$ for general excitationf(t).

$$\mathbf{(2.7)}$$ Verify result with table of particular solution for:
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle f(t)=texp(bt)$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.10)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.10)
 * }
 * }

$$\mathbf{(2.8)}$$ Solve the nonhomogeneous L2-ODE-CC Equation(6.1.1) with the Hyperbolic function:
 * {| style="width:100%" border="0" align="left"

$$ For the coefficients $$(a_0,a_1,a_2)$$, consider two different characteristic equations:
 * $$\displaystyle f(t)=tanht$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.11)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.11)
 * }
 * }

$$\mathbf{(2.8.1)}$$
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle (r+1)(r-2)=0$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.12)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.12)
 * }
 * }

$$\mathbf{(2.8.2)}$$
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle (r-4)^2=0$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.13)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.13)
 * }
 * }

$$\mathbf{(2.9)}$$ For each case in Equation(6.1.12)and Equation(6.1.13), determine the fundamental period of undamped free vibration, and plot the solution for the excitation for about 5 periods, assuming zero initial conditions.

Solution
$$\mathbf{(1)}$$ Multiply Equation(6.1.1) by the integrating factor h(t,y): $$\underbrace{h(t,y)a_2}_{\displaystyle f}y''+\underbrace{h(t,y)a_1y'+h(t,y)a_0y-h(t,y)f(t)}_{\displaystyle g}=0 $$ Recall the 2nd exactness condition for N2-ODEs: $$\displaystyle f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$$ $$\displaystyle f_{xp}+pf_{yp}+2f_y=g_{pp} $$ Calculate the following terms in the exactness condition: $$\displaystyle f_t=a_2h_t\,;\,f_{tt}=a_2h_{tt}\,;\,f_{ty}=a_2h_{ty}\,;\,f_{tp}=0$$ $$\displaystyle f_y=a_2h_y\,;\,f_{yy}=a_2h_{yy}\,;\,f_{yp}=0$$ $$\displaystyle g_t=a_1ph_t+a_0yh_t-f(t)\cdot h_t-h\cdot f'(t)$$ $$\displaystyle g_{tp}=a_1h_t$$ $$\displaystyle g_y=a_1ph_y+a_0yh_y+a_0h-f(t)h_y$$ $$\displaystyle g_{yp}=a_1h_y\,;\,g_p=a_1h\,;\,g_{pp}=0$$ Substituting those calculated terms into 2nd exactness condition: $$\displaystyle a_2h_{tt}+2p\cdot a_2 h_{ty}+p^2\cdot a_2h_{yy}=a_1h_t+\cancel{p\cdot a_1h_y}-\cancel{a_1ph_y}-a_0yh_y-a_0h+f(t)h_y$$ $$\displaystyle 0+p\cdot 0+2\cdot a_2h_y=0$$ From the second equation, we can obtain that $$\displaystyle h_y=0$$ which means that integrating factor $$\displaystyle h(t,y)=h(t)$$. Substituting h(t) into the first equation of 2nd exactness condition, we can obtain an L2-ODE-CC, $$\displaystyle a_2h_{tt}-a_1h_t+a_0h=0$$ Therefore, that PDE reduces into an ODE, which can be solved.

$$\mathbf{(2.1)}$$ Differentiate the RHS of Equation(6.1.3), $$\displaystyle \frac{d}{dt}\left[e^{\alpha t}(\bar a_1y'+\bar a_0y)\right]=\alpha e^{\alpha t}(\bar a_1y'+\bar a_0y)+e^{\alpha t}(\bar a_1 y''+\bar a_0 y')$$ $$\displaystyle =e^{\alpha t}[\bar a_1y''+(\bar a_0+\alpha \bar a_1)y'+\alpha \bar a_0y]$$ Compare the coefficient with the integrand on LHS of Equation(6.1.3), $$\displaystyle e^{\alpha t}[\bar a_1y+(\bar a_0+\alpha \bar a_1)y'+\alpha \bar a_0y]=e^{\alpha t}[a_2y+a_1y'+a_0y]$$ $$\displaystyle \bar a_1=a_2$$ $$\displaystyle \bar a_0+\alpha \bar a_1=a_1$$ $$\displaystyle \alpha \bar a_0=a_0$$

We can find $$\displaystyle (\bar a_1,\ \bar a_0)$$ in terms of $$\displaystyle (a_0,\ a_1,\ a_2)$$ that $$\displaystyle \bar a_1=a_2$$ $$\displaystyle \bar a_0=\frac{a_0}{\alpha}$$

$$\mathbf{(2.2)}$$ Substituting $$\displaystyle \bar a_0$$ and $$\displaystyle \bar a_1$$ into $$\displaystyle \bar a_0+\alpha \bar a_1=a_1$$, we can get that <\br> $$\displaystyle \frac{a_0}{\alpha}+\alpha\cdot a_2=a_1$$ Rearrange it, we obtain the quadratic equation for $$\displaystyle \alpha$$: $$\displaystyle a_2 \alpha^2-a_1\alpha+a_0=0$$

$$\mathbf{(2.3)}$$ Therefore, the reduced-order equation is obtained $$\displaystyle e^{\alpha t}[\bar a_1y'+\bar a_0y]=\int e^{\alpha t}f(t)dt$$ Rewrite it,
 * {| style="width:100%" border="0" align="left"

$$ Equation(6.1.14) is a L1-ODE-CC, which can be solved by the IFM.
 * $$\displaystyle \bar a_1y'+\bar a_0y=e^{-\alpha t}\,\int e^{\alpha t}f(t)dt$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.14)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.14)
 * }
 * }

$$\mathbf{(2.4)}$$ Recall the IFM of general non-homogeneous L1-ODE-VC, $$\displaystyle 1\cdot y'+\underbrace{\frac{Q(x)}{P(x)}}_{a_0(x)}y=\underbrace{\frac{R(x)}{P(x)}}_{b(x)}$$ The integrating factor is $$\displaystyle h(x)=exp[\int^x a_0(s)ds]$$ The solution is $$\displaystyle y(x)=\frac{1}{h(x)}\left[\int^x h(s)b(s)ds+k\right]$$ Divided Equation(6.1.14) by $$\displaystyle \bar a_1$$, $$\displaystyle 1\cdot y'+\underbrace{\frac{\bar a_0}{\bar a_1}}_{\displaystyle \beta}y=\underbrace{\frac{e^{-\alpha t}}{\bar a_1}\, \int e^{\alpha t} f(t)dt}_{\displaystyle b(t)}$$ Therefore, we can obtain the integrating factor $$\displaystyle \bar h(t)=exp[\int\,\frac{\bar a_0}{\bar a_1} dt]$$ $$\displaystyle =exp[\int\,\beta dt]$$ $$\displaystyle =e^{\beta t}$$ Therefore, the solution of Equation(6.1.14) is $$\displaystyle y(t)=\frac{1}{\bar h(t)}\left[ \int^t h(s)b(s)ds+k\right]$$
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle =\frac{1}{e^{\beta t}}\int^t e^{\beta s}\cdot\frac{e^{-\alpha s}}{\bar a_1}\left[\int^s e^{\alpha \tau}f(\tau)d\tau\right]ds$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.15)
 * <p style="text-align:right;">$$\displaystyle (Equation\;6.1.15)
 * }
 * }

$$\mathbf{(2.5)}$$ Substitute $$\displaystyle \bar a_1=a_2\,;\,\bar a_0+\alpha \bar a_1=a_1\,;\,\alpha \bar a_0=a_0$$ into $$\displaystyle \alpha$$ and$$\displaystyle \beta$$, we can obtain that $$\displaystyle \alpha\beta=\alpha \frac{\bar a_0}{\bar a_1}=\frac{a_0}{\bar a_1}=\frac{a_0}{a_2}$$ $$\displaystyle \alpha +\beta= \frac{a_1-\bar a_0}{\bar a_1}+\frac{\bar a_0}{\bar a_1}=\frac{a_1}{\bar a_1}=\frac{a_1}{a_2}$$ Thus, $$\displaystyle (\alpha,\beta)$$ are the roots of the following quadratic equtation: $$\displaystyle (\lambda-\alpha)(\lambda-\beta)=\lambda^2-\underbrace{(\alpha+\beta)}_{\displaystyle \frac{a_1}{a_2}}\lambda+\underbrace{\alpha\beta}_{\displaystyle \frac{a_0}{a_2}}$$ Obviously, it is the same as the quadratic equation obtained by $$\mathbf{(2.2)}$$.

$$\mathbf{(2.6)}$$ Rearrange the solution of Equation(6.1.14) $$\displaystyle \frac{1}{e^{\beta t}}\int^t e^{\beta s}\cdot\frac{e^{-\alpha s}}{\bar a_1}\left[\int^s e^{\alpha \tau}f(\tau)d\tau\right]ds$$ $$\displaystyle =e^{-\beta t}[\int^t\frac{e^{(\beta-\alpha)s}}{\bar a_1}\int^s e^{\alpha \tau}f(\tau)d\tau]ds$$ $$\displaystyle =\frac{e^{-\beta t}}{\bar a_1}[\int^t e^{(\beta-\alpha)s}[\int e^{\alpha s}f(s)ds+k_1]ds]$$ $$\displaystyle =\frac{e^{-beta t}}{\bar a_1}[\int^te^{\beta-\alpha)s}[\int e^{\alpha s}f(s)ds]ds+\int^t e^{(\beta-\alpha)s}k_1ds]$$ $$\displaystyle =\frac{e^{-\beta t}}{\bar a_1}[\int e^{(\beta-\alpha)t}[\int e^{\alpha s}f(s)ds]dt+k_2+\frac{k_1}{\beta-\alpha}e^{(\beta-\alpha)t}]$$ $$\displaystyle =\underbrace{\frac{k_1}{\bar a_1(\beta-\alpha)}e^{-\alpha t}}_{\displaystyle C_1 y_H^1}+\underbrace{\frac{k_2}{\bar a_1}e^{-\beta t}}_{\displaystyle C_2 y_H^2}+\underbrace{\frac{e^{-\beta t}}{\bar a_1}\int e^{(\beta-\alpha)t}[\int e^{\alpha s}f(s)ds]dt}_{\displaystyle y_P}$$ The particular solution $$\displaystyle y_P(t)$$ for general excitation $$\displaystyle f(t)$$ is $$\displaystyle y_P(t)=\frac{e^{-\beta t}}{\bar a_1}\int e^{(\beta-\alpha)t}[\int e^{\alpha s}f(s)ds]dt$$

$$\mathbf{(2.8.1)}$$ $$\displaystyle \alpha$$ and $$\displaystyle \beta$$ is the roots of quadratic equation: $$\displaystyle (r+1)(r-2)=0.$$ Therefore let $$\displaystyle \alpha=-1\,;\,\beta=2$$ As for the excitation f(t) is Hyperbolic function: $$\displaystyle f(t)=tanh(t).$$ For the general solution Equation(6.1.15) $$\displaystyle y(t)=\frac{1}{6a_2}\left[1+2e^{-2t}log(e^{2t}+1)-4e^ttan^{-1}(e^{-t})\right]$$ This is calculated by www.wolframalpha.com.

$$\mathbf{(2.8.2)}$$ The quadratic equation is: $$\displaystyle (r-4)^2=0.$$ Therefore let $$\displaystyle \alpha=\beta=4$$ As for the excitation f(t) is Hyperbolic function: $$\displaystyle f(t)=tanh(t).$$ For the general solution Equation(6.1.15) $$\displaystyle y(t)=\frac{1}{a_2}\left[-\frac{Li_2(-e^{2t}\cdot e^{-4t})}{2}-\frac{e^{-2t}}{2}+\frac{1}{16}\right]$$ where $$\displaystyle Li_2(t)\,$$ is the polylogarithm function. This is calculated by www.wolframalpha.com.