Toronto Math Forum

Compute line integral: $\int_{r} Re(z) \,dz$, where r is line segmemt from 1 to i
$$
f(z) = Re(z)
$$
$$
r(t) = (1-t)z_0 + tz_1 = (1-t)*1 + ti = 1-t+it , 0<= t <= 1
$$
$$
r^{'}(t) = i-1
$$
$$
f(r(t)) = Re(r(t)) = 1-t
$$
$$
\int_{r} f(z) \,dz = \int_{0}^{1}f(r(t))r^{'}(t) \,dt
$$
$$
=\int_{0}^{1}(1-t)(i-1) \,dt
$$
$$
=(i-1)*(t-\tfrac{1}{2} t^2)\Big|_0^1
$$
$$
=(i-1)(1-\tfrac{1}{2}-0)
$$
$$
=\tfrac{1}{2}(i-1)
$$

Here's the solution for sketching.

here's the solution including sketching.

a)Write equation for Wronskian of y1,y2,y3.
$$
W=ce^{-\int p(t)dt}=ce^{-\int 4dt}=ce^{-4t}
$$
b)Find fundamental system of solutions for honogenuous equation, and find Wronskian. Comapare with (a).
$$
r^3+4r^2+r-6=0
$$
$$
(r-1)(r+2)(r+3)=0
$$
$$
r=1, r=-2, r=-3
$$
$$
y_c(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}
$$
$$
W=\begin{vmatrix}
e^t & e^{-2t} & e^{-3t}\\
e^t & -2e^{-2t} & -3e^{-3t}\\
e^t & 4e^{-2t} & 9e^{-3t}\\
\end{vmatrix}
= 12e^{-4t}
$$
similar solution to (a), but $c=12$

(c)Find the general solution.
$$
y_p(t)=Ate^t
$$
$$
y^\prime=Ae^t+Ate^t=Ae^t(1+t)
$$
$$
y^{\prime\prime}=2Ae^t+Ate^t=Ae^t(2+t)
$$
$$
y^{\prime\prime\prime}=3Ae^t+Ate^t=Ae^t(3+t)
$$
$$
y^{\prime\prime\prime}+4y^{\prime\prime}+y^\prime-6y=24e^t
$$
$$
Ae^t(3+t+8+4t+1+t-6t)=24e^t
$$
$$
12Ae^t=24e^t
$$
$$
A=2
$$
Hence, $y(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}+2te^t$

$$
M_{y}=1+6ye^{3x}
$$
$$
N_{x}=4ye^{2x}    
$$
$M_{y} ≠N_{x} $,it is not exact
$$

R_{2} =\frac{ M_{y} -N_{x}}{N}=\frac{1+2ye^{2x}  }{1+2ye^{2x}}=1
$$
$$
μ=e^{∫R_{2}dx} =e^{∫1 dx} =e^x   
$$
Multiplying both sides by $\mu$, we get
$$

ye^x+3y^2e^{3x} +(e^x+2ye^{3x}) y^\prime=0
$$
$$
M_{y}^\prime=e^x+6ye^{3x}
$$
$$
N_{x}^\prime=e^x+6ye^{3x}
$$
$M_{y}^\prime=N_{x}^\prime$,it is exact
$$

∃φ(x,y) such that\ φ_{x} =M^\prime,φ_{y} =N^\prime
$$
$$
φ(x,y)=∫{M^\prime dx}=∫{ye^x+3y^{2}e^{3x}dx}=ye^x+y^{2}e^{3x} +h(y)
$$
$$
φ_{y} =e^x+2ye^{3x} +h(y)^\prime=e^x+2ye^{3x}
$$
Then $h(y)^\prime=0$
$$
$$
Hence h(y)=c
$$

φ(x,y)=ye^x+y^{2}e^{3x} =c
$$
Since y(0)=1
$$
1⋅e^0+1^2⋅e^0=2=c
$$
$$
φ(x,y)=ye^x+y^{2}e^{3x} =2
$$

Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.
$$
\frac{x^2}{y^3}+x(1+y^2\ )\frac{dy}{dx} =0,\mu(x,y)=\frac{1}{xy^3}
$$
First, define $M(x,y)=\ x^2\ y^3$,$N(x,y)=x(1+y^2)$
$$
M_y=\partial y/\partial x\ (x^2\ y^3\ )=3x^2\ y^2
$$
$$
N_y=\partial y/\partial x[x(1+y^2\ )]=1+y^2
$$

Since $M_y\neq N_y$, this implies the given equation is not exact.
Now, multiplying both sides by the integrating factor $\mu(x,y)$, that is
$$
\frac{1}{xy^3}x^2y^3+\frac{1}{xy^3}x(1+y^2\ )\frac{dy}{dx}=x+(\frac{1}{y^3}+\frac{1}{y}) \frac{dy}{dx}=0
$$

Define $M^\prime\ (x,y)=\ x$,$N^\prime\ (x,y)=\frac{1}{y^3}+\frac{1}{y}$
$$
M_y=\partial y/\partial x\ (x)=0
$$
$$
N_y=\partial y/\partial x[\frac{1}{y^3}+\frac{1}{y}]=0
$$

Since $M_y$=$N_y$, this implies the given equation is exact.
Thus, we know there exists a function $\phi(x,y)$ such that $\partial\phi/\partial x=\ M^\prime\ (x,y)=\ x$
Then $\phi(x,y)=\int M^\prime\ dx=\frac{1}{2} x^2+h(y)$
Take derivative on both sides with respect to y we get
$$
\partial\phi/\partial y=h^\prime\ (y)=N^\prime\ (x,y)=\frac{1}{y^3}+\frac{1}{y}
$$

Integrating with respect to y we have
$$
h(y)=\frac{-1}{2}\frac{1}{y^2}+ln|y|+C
$$

Hence, $\phi(x,y)=\frac{1}{2} x^2-\frac{1}{2} \frac{1}{y^2}+ln|y|=C$ is the general solution to the given DE.
Besides, notice that the constant function y(x)=0\ \forall x is also a solution to the given DE.