Toronto Math Forum

Good afternoon,
Please enjoy the last day before reading week.
The question I had on lecture was:
Find the general solution of the given differential equation.
$y''+9y=9sec^23t,  0<t<\pi/6$
For the homogeneous side y''+9y=0:
$r^2+9=0$
$r^2=-9$
$r_1=3i, r_2=-3i$
so the complementary solution is
$y_c(t)=Acos3t+Bsin3t$
For the non-homogeneous side:
$p(t)=0, q(t)=9, g(t)=9sec^23t$ are continuous on $0<t<\pi/6$
$W_{y1,y2}(t)=
\begin{vmatrix}
y_1(t) & y_2(t) \\
y'_1(t) & y'_2(t)
\end{vmatrix}=
\begin{vmatrix}
cos3t & sin3t \\
-3sin3t & 3cos3t
\end{vmatrix}=3$
Thus, $W_1(t)=-\int \frac{y_2(t)g(t)}{W_{1,2}(t)}dt=-\int \frac{sin3t9sec^23t}{3}dt=-3\int \frac{sin3t}{cos^23t}dt$
$=-3\int tan3tsec3tdt=-sec3t$
And, $W_2(t)=\int \frac{y_1(t)g(t)}{W_{1,2}(t)}dt=\int \frac{3cos3t/cos^23t}{3}dt$
$=\int 3sec3t dt=ln|sec3t+tan3t|$
(Be careful the integral of trigonometric here, honestly I made a mistake may lose some mark here on the lecture quiz, we must memorize them)
In conclusion, $Y(t)=y_c(t)+y_p(t)$
$Y(t)=Acos3t+Bsin3t+cos3t(-sec3t)+ln|sec3t+tan3t|sin3t$
$Y(t)=Acos3t+Bsin3t+ln|sec3t+tan3t|sin3t-1$
This is all, hope you have a good reading week!

Hi everyone for quiz 4 we need to get the original equation of
$y''+y'-6y=12e^{-2t}+12e^{3t}$

First, we get the homogeneous form:
$r^2+r-6=0$
$(r+3)(r-2)=0$
$r1=-3, r2=2$

Then we can get:
$y=C_1e^{-2t}+C_2e^{3t}$
$y'=-2C_1e^{-2t}+3C_2e^{3t}$
$y''=4C_1e^{-2t}+9C_2e^{3t}$

so this means:
$-6B-2B+4B=12$
$-6A+3A+9A=12$
Solve to get
$A=2, B=-3$

So we rewrite:
$2e^{3t}-3e^{-2t}$

Plug in to the solution:
$y=2e^{3t}-3e^{-2t}+C_1e^{2t}+C_2e^{-3t}$

Please help correct my answer! Have a good weekend!

Hi everyone, the problem is to prove (x^2)(y^3) + x(1+y^2)y'=0 is not exact, then multiply μ(x,y)=1/(xy^3) to solve the equation

First of all, we observe it follows M(x,y)+N(x,y)∂y/∂x = 0 form, we can get M(x,y)=(x^)(2y^3), N(x,y)= x(1+y^2)=x+xy^2
Then we get My=∂/∂y((x^2)(y^3))=3(x^2)(y^2) and Nx=∂/∂x(x(1+y^2))=1+y^2
As we see My is not equal to Nx, this is not an exact equation.

Second, we try multiplying the integrating factor μ(x,y)=1/(xy^3) on both sides: x+((1/y^3) + (1/y))∂y/dx=0
We do the first step again: M(x,y)=x and N(x,y)=1/y^3 + (1/y)
Derivative them: My=∂/∂y(x) =0 ; Nx=∂/∂x((1/y^3) + (1/y))=0
As we see My=Nx, now this is an exact equation.

Now we need to get the solution: ∫ M∂x+N∂x=(1/2)x^2 - (1/2)y^(-2)+ln|y|=2ln|y|+(x^2)-(1/y^2)=c