User:Egm4313.s12.team15.cynthiah

R1.6.1

Order: 2

Linearity: Linear Superposition : Yes

$$ y''= g = const  \!$$   (1) $$ y''_h=0 \!$$  homogeneous solution (2) $$ y''_p=g \!$$  particular solution (3) By solving we get the following: $$ \bar{y}=y_h + y_p = (y_h + y_p) = 0 +g = g \!$$       (4) $$ \bar{y''}=g \!$$       Since (4) = (1) Superposition is applied (5)

R1.6.2

Order: 1 Linearity: Non-linear

Superposition: No $$ mv'= mg - bv^2  \!$$  (Eq.0) $$ mv' + bv^2 = mg  \!$$  (Eq.0 Rearranged ) $$ mv_h'+b(v_h)^2=0 \!$$   homogeneous solution (Eq. 1 ) $$ mv_p'+b(v_p)^2=mg \!$$  Particular solution (Eq. 2 )

By solving we get the following :

$$\bar{v}^2=v_h+v_p\!$$

$$ m(v_h'+v_p')+ b[(v_h)^2 + (v_p)^2]=mg \!$$

$$m(v_h +v_p)' +b[(v_h)^2 +(v_p)^2]=mg\!$$ $$\bar{v'} + b(v_h+v_p)^2=mg\!$$

$$(v_h+v_p)^2\neq\bar{v}^2 \!$$

Therefore, superposition cannot be applied.

R1.6.3 Order: 1 Linearity:Non- linear

Superposition: No

$$ h'=-k\sqrt{h}\!$$  (Eq.0) $$ h'+k\sqrt{h}=0\!$$  (Eq.0 Rearranged) $${h_h}'+k\sqrt{h_h}= 0 \!$$ Homogeneous solution (Eq. 1 ) $${h_p}'+k\sqrt{h_p}= 0\!$$ Particular solution (Eq. 2)

By solving we get the following:

$$\bar{h}=h_h+h_p\!$$

$${h_h}'+k\sqrt{h_h}+{h_p}'+k\sqrt{h_p}=0\!$$ $$({h_h}'+{h_p}')+k(\sqrt{h_h}+\sqrt{h_p})=0\!$$

$$({h_h}+{h_p}) '+k(\sqrt{h_h}+\sqrt{h_p})=0\!$$

$$(\sqrt{h_h}+\sqrt{h_p})\neq \sqrt\bar{h}\!$$

So superposition cannot be applied.

R1.6.4

Order:2

Linearity: Linear

Superposition: Yes

$$ my''+ky = 0 \!$$  (Eq.0) $$m{y}_h''+ky_h= 0\!$$  Homogeneous solution (Eq. 1 ) $$m{y}_p''+ky_p= 0\!$$   Particular  solution (Eq. 2)

By solving we get the following:

$$\bar{y}=y_h+y_p\!$$ $$\bar{y}=y_h+y_p''\!$$

Summing Eq.1 & Eq. 2 :

$$m({y}_h+{y}_p)''+k(y_h+y_p) = 0\!$$ $$m\bar{y}''+k\bar{y}= 0\!$$

Since:

$$m\bar{y}+k\bar{y}$$ = $$ my+ky \!$$

Then superposition can be applied.

R1.6.5

Order: 2

Linearity: Linear

Superposition: Yes

$$ y'' +w_o^2y=cos wt   \  \ \  \ \  \ \  \ \  \ \  \ \  \  w_o = w \!$$

$$ y_h'' +w_o^2y_h=0\!$$ <p style="text-align:right"> Homogeneous solution (Eq. 1 ) $$ y_p'' +w_o^2y_p=cos wt \!$$  <p style="text-align:right">  Particular  solution (Eq. 2) By solving we obtain the following: $$\bar{y}=y_h+y_p\!$$

$$\bar{y}=y_h+y_p''\!$$ Adding Eq.1 & Eq.2 : $$y_h+w_o^2y_h+y_p+ w_o^2y_p = cos wt\!$$

$$( y_h+y_p)''+w_o^2(y_h+y_p)= cos wt\!$$

$$\bar{y}''+w_o^2\bar{y}= cos wt\!$$

Therefore, superposition can be applied.

R1.6.6

Order: 2

Linearity : Linear

Superposition: Yes

$$ LI'' + RI' + \frac{1}{c}I =E' \!$$

$$ LI_h'' + RI_h' + \frac{1}{c}I_h =0 \!$$<p style="text-align:right"> Homogeneous solution (Eq. 1 )

$$ LI_p'' + RI_p' + \frac{1}{c}I_p =E' \!$$<p style="text-align:right"> Particular  solution (Eq. 2 )

By solving we obtain the following:

$$\bar{I}= (I_h+I_p)\!$$

$$\bar{I}'= (I_h+I_p) '\!$$ $$\bar{I}= (I_h+I_p) \!$$ Adding Eq.1 & Eq. 2:

$$L(I_h+I_p) ''+R(I_h+I_p) ' + \frac{1}{c}(I_h+I_p) = E'\!$$

$$L\bar{I}''+R\bar{I}'+\frac{1}{c}\bar{I}= E'\!$$

Therefore, superposition can be applied.

R1.6.7 Order: 4

Linearity :Linear

Superposition : Yes

$$ EIy^{iv} = f(x) \!$$

$$ y_h= EIy_h^{iv}=0\!$$<p style="text-align:right"> Homogeneous solution (Eq. 1 ) $$ y_p= EIy_p^{iv}=f(x)\!$$<p style="text-align:right"> Particular solution (Eq. 2 )

By solving the following is obtained:

$$\bar{y}=y_h+y_p\!$$

Summing Eq.1 & Eq. 2 :

$$ EIy_h^{iv} + EIy_p^{iv} = f(x)\!$$

$$ EI(y_h + y_p)^{iv}= f(x)\!$$

$$ EI\bar{y}^{iv} = f(x)\!$$ Therefore, superposition can be applied.

R1.6.8

Order: 2

Linearity: Non-linear

Superposition: No

$$ L{\theta }'' + gsin\theta = 0 \!$$ $$ L{\theta_h }'' + gsin\theta_h = 0 \!$$<p style="text-align:right"> Homogeneous solution (Eq. 1 ) $$ L{\theta_p }'' + gsin\theta_p = f(x) \!$$<p style="text-align:right"> Particular solution (Eq. 2 )

By solving the following is obtained:

$$\bar{\theta}= {\theta}_h + {\theta}_p \!$$

Summing Eq. 1 & Eq. 2 :

$$L{\theta_h + \theta_p}''+gsin{\theta_h+\theta_p}=0\!$$ $$\bar{\theta}\neq\ {\theta}_h + {\theta}_p \!$$

Therefore, superposition cannot be applied.

Report 2: 4a)

$$ y{}''+2\pi{}y'+\pi^{2}y=0\!$$

a=2\pi\!

$$ b=\pi^{2}\!$$

$$\lambda ^2+a\lambda +b=0 \! $$

$$\lambda ^2+2\pi\lambda +\pi^{2}=0 \! $$

$$\lambda_1=\frac{1}{2}-2\pi+\sqrt{(-2\pi)^2-4(\pi^{2})}=-\pi\!$$

$$\lambda_2=\frac{1}{2}-2\pi-\sqrt{(-2\pi)^2-4(\pi^{2})}=-\pi\!$$

Both solutions above show the double root characteristic so the following is applied:

$$y_1=e^{-\pi{x}}\!$$

$$y_2=uy_1\!$$

$$ u=c_1x+c_2\!$$  This is due to double integration.

So, $$y_2=(c_1x+c_2)y_1\!$$

Choose $$ c_1= 1 c_2=0\!$$

$$y_2= (x+0)y_1\!$$

$$y_2=xy_1\!$$

$$ y_2=xe^{-\pi{x}}\!$$

$$ y=(c_1+c_2x)e^{-\pi{x}}\!$$ General Solution

Check by substitution

$$ y=(c_1+c_2x)e^{-\pi{x}}\!$$

$$ y'=-\pi{}c_1e^{-\pi{x}}+c_2e^{-\pi{x}}-\pi{}c_2e^{-\pi{x}}\!$$

$$ y''=\pi^{2}c_1e^{-\pi{x}}-\pi{}c_2e^{-\pi{x}}+\pi^{2}c_2xe^{-\pi{x}}-\pi{}c_2e^{-\pi{x}}\!$$

$$\pi^{2}c_1e^{-\pi{x}}-\pi{}c_2e^{-\pi{x}}+\pi^{2}c_2xe^{-\pi{x}}-\pi{}c_2e^{-\pi{x}}+ 2\pi{}(-\pi{}c_1e^{-\pi{x}}+c_2e^{-\pi{x}}-\pi{}c_2e^{-\pi{x}})+ \pi^{2}(e^{-\pi{x}}c_1+e^{-\pi{x}}xc_2)=0 \!$$

Since all terms cancel out then by substitution the general solution does hold true.

4b) $$ 10y{}''-32{}y'+25.6{}y=0\!$$

Divide by 10 to simplify equation :

$$ y{}''-3.2{}y'+2.56{}y=0\!$$

$$ a=-3.2\!$$

$$ b=2.56\!$$

$$ \lambda ^2+a\lambda +b=0 \! $$

$$ \lambda ^2-3.2\lambda +2.56=0 \! $$

Solving for the roots:

$$ \lambda_1=\frac{1}{2}-(-3.2)+\sqrt{(-3.2)^2-4(2.56)}=1.6\!$$

$$ \lambda_2=\frac{1}{2}-(-3.2)-\sqrt{(-3.2)^2-4(2.56)}=1.6\!$$

$$\lambda_1=\lambda_2\!$$ This shows the double root characteristic.

So, the following is applied

$$ y_1=e^{1.6x}\!$$

$$y_2=uy_1\!$$

$$ u=c_1x+c_2\!$$  This is due to double integration

So

$$y_2=(c_1x+c_2)y_1\!$$

Choose $$ c_1= 1 c_2=0\!$$

$$y_2= (x+0)y_1\!$$

$$y_2=xy_1\!$$

$$ y_2=xe^{1.6x}\!$$

$$ y=(c_1+c_2x)e^{1.6x}\!$$  General Solution

 Check by Substitution

$$ y''=2.56c_1e^{1.6x}+1.6c_2e^{1.6x}+1.6c_2xe^{1.6x}+2.56c_2xe^{1.6x}\!$$

$$y'=1.6c_1e^{1.6x}+c_2e^{1.6x}+ 1.6c_2xe^{1.6x} \!$$

$$y=c_1e^{1.6x}+c_2xe^{1.6x} \!$$

$$10(2.56c_1e^{1.6x}+1.6c_2e^{1.6x}+1.6c_2xe^{1.6x}+2.56c_2xe^{1.6x})-32(1.6c_1e^{1.6x}+c_2e^{1.6x}+ 1.6c_2xe^{1.6x})+25.6(c_1e^{1.6x}+c_2xe^{1.6x})=0\!$$

Since all terms cancel out then through substitution it can be confirmed that the general solution

$$ y=(c_1+c_2x)e^{1.6x}\!$$ does hold true.