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1

Term Test 1 / Re: Problem 2 (morning)

« on: October 23, 2019, 10:05:07 AM »

Question 2:
$
xy''-(2x+1)y'+(x+1)y=0
$

Find wronskian form and $y_2$, given $y_1(x)=e^x$.
$y(1)=0,y'(1)=e$

Solution:
$y''-\frac{2x+1}{x}y'+\frac{x+1}{x}y=0$
$w=ce^{-\int p(x)\mathrm{d}x}=ce^{\int \frac{2x+1}{x}\mathrm{d}x}=ce^{2x}e^{\ln x}$

Let $c=1$, $w=xe^{2x}$
$
\begin{bmatrix}
y_1&y_2\\
y_1'&y_2'\\
\end{bmatrix}
=
\begin{bmatrix}
e^x&y_2\\
e^x&y_2'\\
\end{bmatrix}
=
e^xy_2'-e^xy_2=xe^{2x}
$
$
y_2'-y_2=xe^x
$
$
M=e^{\int-1\mathrm{d}x }=e^{-x}
$

Multiple $e^{-x}$ on both sides.
$
e^{-x}y_2'-e^{-x}y_2=x
$
$(e^{-x}y_2)'=x$
$
y_2=\frac12x^2e^x+e^x
$
$
y=c_1y_1+c_2y_2=c_1e^x+c_2\left(\frac12x^2e^x+e^x\right)
$

plug in $y(1)=0,y'(1)=e$.

we have $y=-\frac12e^x+\frac12x^2e^x$.

2

Term Test 1 / Re: Problem 3 (morning)

« on: October 23, 2019, 10:02:25 AM »

Question 3:
$
y''-6y'+8y=48\sin h (2x), y(0)=0, y'(0)=0.
$

Solution:
$
r^2-6r+8=0
$
$r=4, r=2$
$y_c(x)=c_1e^{4t}+c_2e^{2t}$
$48\sin h (2x)=24\cdot e^{2x}-24\cdot e^{-2x}$
$
y_p(x)=Axe^{2x}, y_p'=2Axe^{2x}+Ae^{2x},
$
$
y_p''(x)=4Axe^{2x}+2Ae^{2x}
$
$
y''-6y'+8y=-2Ae^{2x}=24e^{2x}
$
$
A=-12,y_p(x)=-12xe^{2x}
$
$
y_p(x)=Be^{-2x},y_p'(x)=-2Be^{-2x},y_p''(x)=4Be^{-2x}
$
$
y''-6y'+8y=24Be^{-2x}=-24e^{-2x}
$
$
B=-1,y_p(x)=-e^{-2x}
$
$
y=y_c+y_p=c_1e^{4t}+c_2e^{2t}-12xe^{2x}-e^{-2x}
$

Plug in $y(0)=0$, $y'(0)=0$.

we have $c_1=4,c_2=3$
$
y=4e^{4t}+3e^{2t}-12xe^{2x}-e^{-2x}
$

3

Term Test 1 / Re: Problem 4 (morning)

« on: October 23, 2019, 10:01:09 AM »

Question 4:
$
y''-6y'+25y=16e^{3x}+102\sin (x), y(0)=0, y'(0)=0
$

Solution:
$
\Gamma^2-6\Gamma+25=0, \Gamma=3\pm 4\pi
$
$
y_c(x)=c_1e^{3x}\cos (4x)+c_2e^{3x}\sin (4x)
$
$
y_p(x)=Ae^{3x},y_p'(x)=3Ae^{3x},y_p''(x)=9Ae^{3x}
$
$
y''-6y'+25y=16Ae^{3x}=16e^{3x}
$
$
A=1,y_p(x)=e^{3x}
$
$
y_p(x)=B\cos (x)+C\sin x, y_p'(x)=-B\sin x+C\cos x,
$
$
y_p''(x)=-B\cos x-C\sin x
$
$
y''-6y'+25y=102\sin x
$
$
B=1,C=4,y_p(x)=\cos x+4\sin x
$
$
y=y_c+y_p=c_1e^{3x}\cos (4x) +c_2e^{3x}\sin (4x)+e^{3x}+\cos x+4\sin x
$

Plug in $y(0)=0,y'(0)=0$

we have $c_1=-2,c_2=-\frac14$
$
y=-2e^{3x}\cos (4x)-\frac14e^{3x}\sin (4x)+e^{3x}+\cos x+4\sin x
$

4

Term Test 1 / Re: Problem 1 (morning)

« on: October 23, 2019, 09:57:47 AM »

Question 1:
$
[-y\sin(x)+y^3\cos(x)]+[3\cos(x)+5y^2\sin(x)]y'=0,
y\left(\frac{\pi}{4}\right)=\sqrt2.
$

Solution:
Since $M_y$ does not equal to $N_x$, the equation is not exact.
$
R_1=\frac{M_y-N_x}{M}=\frac{2(\sin x-y^2\cos x)}{-y(\sin x-y^2\cos x)}=-\frac{2}{y}.
$
$
M=e^{-\int R_1\mathrm{d}y}=e^{\int\frac{2}{y}\mathrm{d}y}=e^{2\ln y}=y^2.
$

Multiple both side with $y^2$.
$
[-y^3\sin (x)+y^5\cos(x)]+[3y^2\cos(x)+5y^4\sin(x)]y'=0
$
$
\varphi_x=M
$
$
\varphi=\int M \mathrm{d}x=\int -y^3\sin x +y^5\cos x \mathrm{d}x=y^3\cos x+y^5\sin x +h(y)
$
$
\varphi_y=N=3y^2\cos(x)+5y^4\sin (x)+h'(y)
$
$
h(y)=C
$
$
\varphi (x,y)=y^3\cos x+y^5\sin x=C
$
$
C=\sqrt2^3x\cos \frac{\pi}{4}+\sqrt2^5x\sin \frac{\pi}4=6
$
$
\varphi(x,y)=y^3\cos x+y^5\sin x=6
$