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##### Quiz-5 / TUT 5103 Quiz5
« on: November 01, 2019, 02:00:00 PM »
\noindent Verify that the given functions y 1and y 2 satisfy the corresponding homogeneous equation;
then find a particular solution of the given nonhomogeneous equation.
$$\begin{array}{l}{(1-t) y^{\prime \prime}+t y^{\prime}-y=2(t-1)^{2} e^{-t}, 0<t<1} \\ {y_{1}(t)=e^{t}, y_{2}(t)=t}\end{array}$$

$$y^{\prime \prime}+\frac{t}{1-t} y^{\prime}-\frac{1}{1-t} y=-2(t-1) e^{-t}$$
$$\begin{array}{l}{w=\left|\begin{array}{ll}{e^{t}} & {t} \\ {e^{t}} & {1}\end{array}\right|=e^{t}-t e^{t}} \\ {w_{1}=\left|\begin{array}{ll}{0} & {t} \\ {1} & {1}\end{array}\right|=-t \quad w_{2}=\left|\begin{array}{ll}{e^{t}} & {0} \\ {e^{t}} & {1}\end{array}\right|=e^{\tau}}\end{array}$$
\begin{aligned} Y(t) &=e^{t} \int \frac{-t \cdot\left(-2(t-1) e^{-t}\right)}{e^{t}-t e^{t}} d t+t \int \frac{e^{t}\left(-2(t-1) e^{-t}\right)}{e^{t}-t e^{t}} d t \\ &=e^{t} \int \frac{-t \cdot\left(t^{2}(t-1) e^{-t}\right)}{e^{t}(1-t)}+t \int \frac{e^{t}\left(t+2(t-1) e^{-t}\right)}{e^{t}(1-t)} d t \\ &=e^{t} \int-2 t e^{-2 t} d t+\int 2 e^{-t} d t \\ &=-2 e^{t} \int t e^{-2 t} d t-2 t e^{-t} \end{aligned}
$$\begin{array}{ll}{u=t} & {d v=e^{-2 t}} \\ {d u=d t} & {v=-\frac{1}{2} e^{-2 t}}\end{array}$$
$$\begin{array}{l}{\displaystyle =-2 e^{t}\left(-\frac{1}{2} e^{-2 t} \cdot t+\int \frac{1}{2} e^{-2 t} d t\right)+t \int 2 e^{-t} d t} \\ {\displaystyle =-2 e^{t}\left(-\frac{1}{2} e^{-2 t} t-\frac{1}{4} e^{-2 t}\right)-2 t e^{-t}} \\ {\displaystyle =t e^{-t}+\frac{1}{2} e^{-t}-2 t e^{-t}} \\ {\displaystyle =\left(\frac{1}{2}-t\right) e^{-t}}\end{array}$$

$\therefore$ the particular solution of the given non-homogeneous equation is
$$Y(t)=\left(\frac{1}{2}-t\right) e^{-t}$$

2
##### Term Test 1 / Re: Problem 2 (afternoon)
« on: October 23, 2019, 03:15:06 PM »
$$(2 x+1) x y^{\prime \prime}+(2 x+2) y^{\prime}-2 y=0$$

Find $w,$ and $y_{1}=x+1$

Find solution when $y(-1)=1 \quad y^{\prime}(-1)=0$
$$y^{\prime \prime}+\frac{2 x+2}{(2 x+1) x} y^{\prime}-\frac{2}{(2 x+1) x} y=0$$
$$w=c e^{-\int \frac{2 x+2}{(2 x+1) x}} d x$$
$$\begin{array}{l}{\int \frac{2 x+2}{(2 x+1) x} d x} \\ {=\quad 2 \int \frac{x+1}{x(2 x+1)} \quad d x}\end{array}$$
\begin{aligned} \frac{x+1}{x(2 x+1)} &=\frac{A}{x}+\frac{B}{2 x+1} \\ x+1=& A(2 x+1)+B x \\ x+1=& A 2 x+A+B x \\ x+1=& x(2 A+B)+A \end{aligned}
$$A=1 \quad B=-1$$
\begin{aligned} \frac{x+1}{x(2 x+1)}=& \frac{1}{x}-\frac{1}{2 x+1} \\ \int \frac{x+1}{x(2 x+1)} &=\int \frac{1}{x} d x-\int \frac{1}{2 x+1} d x \\ &=\ln |x |-\int \frac{1}{2 x+1} \end{aligned}
$$\int \frac{1}{2 x+1} d x$$
\begin{aligned} & u=2 x+1 \quad \frac{d u}{d x}=2 \\=& \int \frac{1}{u} \cdot \frac{d u}{2} \\=& \frac{1}{2} \ln |u| \\=& \frac{1}{2} \ln |2 x+1| \end{aligned}
\begin{aligned} 2 \int \frac{x+1}{(2 x+1) x}=2 \ln |x| &-\ln |2 x+1| \\ W=\operatorname{ce}^{-\int \frac{2 x+12}{(2 x+1) x}} &=c e^{-2 \ln x+\ln 2 x+1} \\ &=c e^{-2 \ln x} \cdot e^{\ln 2 x+1} \\ &=c \frac{1}{x^{2}} \cdot 2 x+1 \end{aligned}
Let c=1$$=\frac{2 x+1}{x^{2}}$$ $$w=\left|\begin{array}{ll}{y_{1}} & {y_{2}} \\ {y_{1}^{\prime}} & {y_{2}^{\prime}}\end{array}\right|=\left|\begin{array}{cc}{x+1} & {y_{2}} \\ {1} & {y_{2}^{\prime}}\end{array}\right|=\frac{2 x+1}{x^{2}}$$ $$\begin{array}{l}{(x+1) y_{2}^{\prime}-y_{2}=\frac{2 x+1}{x^{2}}} \\ {y_{2}^{\prime}-\frac{1}{x+1} y_{2}=\frac{2 x+1}{x^{2}(x+1)}}\end{array}$$ $$\begin{array}{l}{u=e^{\int p(x) d x}=e^{-\int \frac{1}{x+1}}=e^{-\ln (x+1)}=\frac{1}{x+1}} \\ {\frac{1}{x+1} y_{2}^{\prime}-\frac{1}{(x+1)^{2}} y_{2}=\frac{2 x+1}{x^{2}(x+1)^{2}}}\end{array}$$ $$\begin{array}{l}{\left(\frac{1}{x+1} y_{2}\right)^{\prime}=\frac{2 x+1}{x^{2}(x+1)^{2}}} \\ {\frac{1}{x+1} y_{2}=\int \frac{2 x+1}{x^{2}(x+1)^{2}}}\end{array}$$ $$\begin{array}{l}{\int \frac{2 x+1}{x^{2}(x+1)^{2}} d x} \\ {\quad u=x(x+1) \frac{d u}{d x}=2 x+1} \\ {=\int \frac{1}{u^{2}} d u} \\ {=-\frac{1}{u}} \\ {=-\frac{1}{x(x+1)}+c}\end{array}$$ \begin{aligned} \frac{1}{x+1} y_{2} &=-\frac{1}{x (x+1)}+i \\ y_{2}=&-\frac{1}{x}+c(x+1) \\ &=-\frac{1}{x}+(x+1) \end{aligned} \quad \text { let } c=1 \begin{aligned} y=c_{1} y_{1}+c_{2} y_{2} &\left.=c_1({x}+1\right)+c_{2}\left(-\frac{1}{x}+(x+1)\right) \\ &=c_{1} x+c_{1}-\frac{c_{2}}{x}+c_{2} x+c_{2} \end{aligned} $$\begin{array}{l}{\qquad \begin{array}{l}{y(-1)=1} \\ {1=-c_{1}+c_{1}+c_{2}-c_{2}+c_{2}} \\ {1=c_{1}} \\ {y^{\prime}=c_{1}+\frac{c_{2}}{x^{2}}+c_{2}}\end{array}}\end{array}$$ \begin{aligned} y^{\prime}(-1)&=0 \\ 0 &=c_{1}+c_{2}+c_{2} \\ &=c_{1}+2 c_{2} \end{aligned} $$\left\{\begin{array}{l}{c_{2}=1} \\ {c_{1}+2 c_{2}=0}\end{array}\right.$$ $$c_{1}=-2$$ \begin{aligned} y &=-2 x-2-\frac{1}{x}+x+1 \\ &=-x-\frac{1}{x}-1 \end{aligned} 3 ##### Quiz-4 / TUT 5103 Quiz4 « on: October 18, 2019, 02:00:02 PM » $$\begin{array}{l}{y^{\prime \prime}+y=0, \quad y\left(\frac{\pi}{3}\right)=2, \quad y^{\prime}\left(\frac{\pi}{3}\right)=-4} \\ {r^{2}+1=0}\end{array}$$ \begin{aligned} r &=\frac{-0 \pm \sqrt{0-4 \cdot 1} \cdot 1}{2} \\ &=\frac{-0+2 i}{2} \\ &=0 \pm i \end{aligned} $$\begin{array}{l}{y=c_{1} e^{o t} \cos t+c_{2} e^{o t} \sin t} \\ {y=c_{1} \cos t+c_{2} \sin t} \\ {y^{\prime}=-c_{1} \sin t+c_{2} \cos t}\end{array}$$ $$\left\{\begin{array}{l}{2=\frac{1}{2} c_{1}+\frac{\sqrt{3}}{2} c_{2}} \\ {-4=-\frac{\sqrt{3}}{2} c_{1}+\frac{1}{2} c_{2}}\end{array}\right.$$ \begin{aligned} c_{1} &=1+2 \sqrt{3}\quad c_{2}=\sqrt{3}-2 \\ \therefore y &=(1+2 \sqrt{3}) \cos t+(\sqrt{3}-2) \sin t \end{aligned} 4 ##### Quiz-3 / MAT244TUT5103 Quiz3 « on: October 11, 2019, 02:00:03 PM » Find the Wronskian of the given pair of functions. $$x, x e^{x}$$ $$\begin{array}{l}{(x)^{\prime}=1} \\ {\left(x e^{x}\right)^{\prime}=x e^{x}+e^{x}}\end{array}$$ $$W=\left|\begin{array}{cc}{x} & {x e^{x}} \\ {1} & {x e^{x}+e^{x}}\end{array}\right|=x^{2} e^{x}+x e^{x}-x e^{x}=x^{2} e^{x}$$ 5 ##### Quiz-2 / MAT24f4 TUT5103 Quiz2 « on: October 04, 2019, 02:02:30 PM » $$(ye^{2xy}+x)+bxe^{2xy}y^{\prime}=0.$$ \begin{aligned} M&=ye^{2xy}+x\\ M_y&=e^{2xy}+2xye^{2xy}\\ N_x&=bxe^{2xy}+2bxye^{2xy}\\ e^{2xy}&+2xye^{2xy}=bxe^{2xy}+2bxye^{2xy}\\ &b=1\\ \end{aligned} Substitute(ye^{2xy}+x)+xe^{2xy}y^{\prime}=0\begin{aligned} M&=ye^{2xy}+x\qquad N=xe^{2xy}\\ M_y&=N_x\qquad \therefore \text{the function now is exact.}\\ \varphi_x&=M\\ \varphi&=\int ye^{2xy}+xdx\\ &=\frac{ye^{2xy}}{2y}+\frac{x^2}{2}+h(y)=\frac{e^{2xy}}{2}\frac{x^2}{2}+h(y)\\ \varphi_y&=\frac{2xe^{2xy}}{2}+h^{\prime}(y)\\ &=xe^{2xy}+h^{\prime}(y)\\ h^{\prime}(y)&=0\\ h(y)&=\text{constant}\\ \varphi&=\frac{e^{2xy}}{2}+\frac{x^2}{2}=c \end{aligned} 6 ##### Quiz-2 / MAT24f4 TUT5103 Quiz2 « on: October 04, 2019, 02:01:16 PM » $$(ye^{2xy}+x)+bxe^{2xy}y^{\prime}=0.$$ \begin{aligned} M&=ye^{2xy}+x\\ M_y&=e^{2xy}+2xye^{2xy}\\ N_x&=bxe^{2xy}+2bxye^{2xy}\\ e^{2xy}&+2xye^{2xy}=bxe^{2xy}+2bxye^{2xy}\\ &b=1\\ \end{aligned} Substitute(ye^{2xy}+x)+xe^{2xy}y^{\prime}=0\begin{aligned} M&=ye^{2xy}+x\qquad N=xe^{2xy}\\ M_y&=N_x\qquad \therefore \text{the function now is exact.}\\ \varphi_x&=M\\ \varphi&=\int ye^{2xy}+xdx\\ &=\frac{ye^{2xy}}{2y}+\frac{x^2}{2}+h(y)=\frac{e^{2xy}}{2}\frac{x^2}{2}+h(y)\\ \varphi_y&=\frac{2xe^{2xy}}{2}+h^{\prime}(y)\\ &=xe^{2xy}+h^{\prime}(y)\\ h^{\prime}(y)&=0\\ h(y)&=\text{constant}\\ \varphi&=\frac{e^{2xy}}{2}+\frac{x^2}{2}=c \end{aligned} 7 ##### Quiz-2 / MAT24f4 TUT5103 Quiz2 « on: October 04, 2019, 02:00:45 PM » $$(xy^{2}+bx^{2}y)+(x+y)x^2y^{\prime}=0$$ \begin{aligned} M&=xy^2+bx^2y\qquad N=(x+y)x^{2}=x^3+yx^2\\ M_y&=2yx+bx^2\qquad N_x=3x^2+2yx\\ \therefore &b=3 \end{aligned} $$(xy^{2}+3x^{2}y)+(x+y)x^2y^{\prime}=0$$ \begin{aligned} M&=xy^2+3x^2y\qquad N=(x+y)x^{2}=x^3+yx^2\\ M_y&=2yx+3x^2\qquad N_x=3x^2+2yx\\ \therefore &M_y=N_x\quad\therefore \text{the function is exact.} \end{aligned} \begin{aligned} \exists ~\text{s.t.} \varphi_x&=M\\ \varphi&=\int M d x\\ &=\int xy^2+3x^2ydx\\ &=\frac{x^2y^2}{2}+x^3y+h(y)\\ \varphi_y&=\frac{2yx^2}{2}+x^3+h^{\prime}(y)\\ &=x^2y+x^3=h^{\prime}(y)\\ &=x^3+x^2y\\ h^{\prime}(y)&=0\\ h(y)&=\text{constant}\\ \varphi&=\frac{x^2y^2}{2}+x^3y=c \end{aligned} 8 ##### Quiz-2 / MAT24f4 TUT5103 Quiz2 « on: October 04, 2019, 02:00:11 PM » $$(ye^{2xy}+x)+bxe^{2xy}y^{\prime}=0.$$ \begin{aligned} M&=ye^{2xy}+x\\ M_y&=e^{2xy}+2xye^{2xy}\\ N_x&=bxe^{2xy}+2bxye^{2xy}\\ e^{2xy}&+2xye^{2xy}=bxe^{2xy}+2bxye^{2xy}\\ &b=1\\ \end{aligned} Substitute(ye^{2xy}+x)+xe^{2xy}y^{\prime}=0\$

\begin{aligned} M&=ye^{2xy}+x\qquad N=xe^{2xy}\\ M_y&=N_x\qquad \therefore \text{the function now is exact.}\\ \varphi_x&=M\\ \varphi&=\int ye^{2xy}+xdx\\ &=\frac{ye^{2xy}}{2y}+\frac{x^2}{2}+h(y)=\frac{e^{2xy}}{2}\frac{x^2}{2}+h(y)\\ \varphi_y&=\frac{2xe^{2xy}}{2}+h^{\prime}(y)\\ &=xe^{2xy}+h^{\prime}(y)\\ h^{\prime}(y)&=0\\ h(y)&=\text{constant}\\ \varphi&=\frac{e^{2xy}}{2}+\frac{x^2}{2}=c \end{aligned}

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