Toronto Math Forum

$$f(z)=\frac{8}{(z-3)(z+5)}=\frac{1}{z-3}-\frac{1}{z+5}$$
(a)as$|z|<3$,
$$f(z)= -\frac{1}{3} \frac{1}{1-\frac{z}{3}}- \frac{1}{5} \frac{1}{1-(-\frac{z}{5})}=-\frac{1}{3}\sum_{n=0}^{\infty}(\frac{z}{3})^n-\frac{1}{5}\sum_{n=0}^{\infty}(-\frac{z}{5})^n$$
(b)as$3<|z|<5$,
$$f(z)= \frac{1}{z} \frac{1}{1-\frac{3}{z}}- \frac{1}{5} \frac{1}{1-(-\frac{z}{5})}=\frac{1}{z}\sum_{n=0}^{\infty}(\frac{3}{z})^n-\frac{1}{5}\sum_{n=0}^{\infty}(-\frac{z}{5})^n=\sum_{n=0}^{\infty}3^nz^{-n-1}-\frac{1}{5}\sum_{n=0}^{\infty}(-\frac{z}{5})^n$$
(c)as$|z|>5$,
$$f(z)= \frac{1}{z} \frac{1}{1-\frac{3}{z}}- \frac{1}{z} \frac{1}{1-(-\frac{5}{z})}=\sum_{n=0}^{\infty}3^nz^{-n-1}-\sum_{n=0}^{\infty}(-5)^nz^{-n-1}=\sum_{n=0}^{\infty}z^{-n-1}(3^n-(-5)^n)$$

$$f(z)=\frac{16}{(z^2-16)(z^2+25)}$$
By doing partial fraction, we can get
$$f(z)=\frac{2}{41}\frac{1}{z-1}-\frac{2}{41}\frac{1}{z+4}-\frac{16}{41}\frac{1}{10i}(\frac{1}{z-5i}-\frac{1}{z+5i})$$
a) as $|z|<4,$
$$f(z)=\frac{2}{41}\frac{(-1)}{4}\frac{1}{1-\frac{z}{4}}-\frac{2}{41}\frac{1}{4}\frac{1}{1-\frac{(-z)}{4}}+\frac{16}{41}\frac{1}{10i}\frac{1}{5i}(\frac{1}{1-\frac{z}{5i}}+\frac{1}{1-\frac{-z}{5i}})$$
$$∴f(z)=\frac{-2}{164}(\sum_{n=0}^{\infty}(\frac{z}{4})^n+\sum_{n=0}^{\infty}(\frac{-z}{4})^n)-\frac{16}{2050}(\sum_{n=0}^{\infty}(\frac{z}{5i})^n+\sum_{n=0}^{\infty}(\frac{-z}{5i})^n)$$
Similarily,
b)as $<4|z|<5$,
$$f(z)=\frac{2}{41z}(\sum_{n=0}^{\infty}(\frac{4}{z})^n-\sum_{n=0}^{\infty}(\frac{-4}{z})^n)-\frac{16}{2050}(\sum_{n=0}^{\infty}(\frac{z}{5i})^n+\sum_{n=0}^{\infty}(\frac{-z}{5i})^n)$$
c)as $|z|>5$,
$$f(z)=\frac{2}{41z}(\sum_{n=0}^{\infty}(\frac{4}{z})^n-\sum_{n=0}^{\infty}(\frac{-4}{z})^n)-\frac{16}{410iz}(\sum_{n=0}^{\infty}(\frac{5i}{z})^n-\sum_{n=0}^{\infty}(\frac{-5i}{z})^n)$$

There is a mistake in your answer. Plugging in $y(0)=0$ , you should get $0=C_1+C_2-\frac{1}{2}$ . Therefore, $C_1=\frac{1}{2}, C_2=0$
$$∴y(t)=\frac{1}{2}e^{-7t}-\frac{1}{2}e^t+4te^{-t}$$

First, we consider $y''+6y'+10y=0​$
$$r^2+6r+10=0$$
$$∴r=-3\pm i​$$
$$∴y_c(t)=C_1e^{-3t}\cos t+C_2e^{-3t}\sin t$$
Second, we consider $y''+6y'+10y=5e^{-3t}​$, let $$y_{p1}(t)= Ae^{-3t}, y'_{p1}(t)=-3Ae^{-3t}, y''_{p1}(t)=9Ae^{-t}​$$
$$9Ae^{-3t} -18Ae^{-3t}+10Ae^{-3t} =5e^{-3t}$$
$$∴A=5$$
$$∴y_{p1}(t)= 5e^{-3t}$$
Third, we consider $y''+6y'+10y=13\cos t$ , let $$y_{p2}(t)= B\cos t+C\sin t, y'_{p2}(t)= -B\sin t+C\cos t,y''_{p2}(t)= -B\cos t-C\sin t​$$
$$∴-B\cos t-C\sin t+(-B\sin t+C\cos t)+10(B\cos t+C\sin t)=13\cos t$$
$$∴9C-6B=0, 9B+6C=13 $$
$$∴ B=1, C=\frac{2}{3}$$
$$∴y_{p2}(t)= \cos t+\frac{2}{3}\sin t​$$
Therefore,
$$y(t)=y_c(t)+y_{p1}(t)+y_{p2}(t) =C_1e^{-3t}\cos t+C_2e^{-3t}\sin t+5e^{-3t}+\cos t+\frac{2}{3}\sin t$$

$$M=y(4lnx + 4lny +1)=y(4ln(xy)+1),N=x(lnx+lny+1)=x(ln(xy)+1)$$
$$M_y =4ln(xy) +5, N_x=ln(xy)+2$$
Since $M_y\ne N_x​$​, it is not exact.
$$R=\frac{M_y-N_x}{N}=\frac{3ln(xy)+3}{x(ln(xy)+1)} = \frac{3(ln(xy)+1)}{x(ln(xy)+1)}=\frac{3}{x}$$
R is a function of x only, so there is an integrating factor of the form
$$\mu=e^{\int R dx}=e^{\int \frac{3}{x} dx}=e^{3lnx }=x^3$$
multiply the differential equation by $\mu​$
$$x^3y(4ln(xy) +1)+x^4(ln(xy)+1)y'=0$$
Since ​$M_y = N_x​$, It is exact now. There exist a ​ $\psi(x,y)​$ such that ​$\psi_x=M, \psi_y=N​$ .
$$\psi=\int M dx=\int x^3y(4ln(xy) +1)dx=x^4yln(xy)+h(y)$$
$$\psi_y=x^4ln(xy)+x^4+h'(y)=x^4(ln(xy)+1)=N$$
$$∴h'(y)=0, h(y)=C$$
$$Therefore, \psi(x,y)= x^4yln(xy)=C$$
as x=1, y=1, we get C=0
$$∴x^4yln(xy)=0$$