1. ## Ode

My Lecturer give me this problems, help please, ...not all, .. one number is all right, at least I'll get an example to solve the other ....

$
\begin{gathered}
1).\frac{{dy}}
{{dx}} + 3\frac{y}
{x} = 6x^2 \hfill \\
2).((\cos ^2 x - y\cos x))dx - (1 + \sin x)dy = 0 \hfill \\
3).x^2 \frac{{dy}}
{{dx}} + xy = xy^3 \hfill \\
4).\frac{{d^2 y}}
{{dx^2 }} - 2\frac{{dy}}
{{dx^{} }} - 3y = 2e^x - 10\sin x \hfill \\
5).\frac{{d^2 y}}
{{dx^2 }} + y = \tan x\sec x \hfill \\
\end{gathered}

$

Thank you very much

2. Originally Posted by Singular

$
\begin{gathered}
1).\frac{{dy}}
{{dx}} + 3\frac{y}
{x} = 6x^2 \hfill \\
2).((\cos ^2 x - y\cos x))dx - (1 + \sin x)dy = 0 \hfill \\
3).x^2 \frac{{dy}}
{{dx}} + xy = xy^3 \hfill \\
4).\frac{{d^2 y}}
{{dx^2 }} - 2\frac{{dy}}
{{dx^{} }} - 3y = 2e^x - 10\sin x \hfill \\
5).\frac{{d^2 y}}
{{dx^2 }} + y = \tan x\sec x \hfill \\
\end{gathered}

$
1)Has form $y'+p(x)y = q(x)$ so use integrating factor.
2)Has form $p(x,y)+q(x,y)y' = 0$ with $\partial_y p(x,y) = \partial_x q(x,y)$.
3)Divide by $x^2$ to bring to form as in #1.

3. Originally Posted by Singular
My Lecturer give me this problems, help please, ...not all, .. one number is all right, at least I'll get an example to solve the other ....

$
\begin{gathered}
1).\frac{{dy}}
{{dx}} + 3\frac{y}
{x} = 6x^2 \hfill \\
2).((\cos ^2 x - y\cos x))dx - (1 + \sin x)dy = 0 \hfill \\
3).x^2 \frac{{dy}}
{{dx}} + xy = xy^3 \hfill \\
4).\frac{{d^2 y}}
{{dx^2 }} - 2\frac{{dy}}
{{dx^{} }} - 3y = 2e^x - 10\sin x \hfill \\
5).\frac{{d^2 y}}
{{dx^2 }} + y = \tan x\sec x \hfill \\
\end{gathered}

$

Thank you very much
4) is a second order DE with constant coefficients. Get the homogeneous solution and use the method of undetermined coefficients to construct a particular solution.

5) Variation of parameters method might work.

4. Hello, Singular!

Some help . . .

$1)\;\;\frac{dy}{dx} + 3\,\frac{y}{x} \:= \:6x^2$
We have: . $\frac{dy}{dx}+ \frac{3}{x}\,y \;=\;6x^2$

Integrating factor: . $I \;=\;e^{\int\frac{3}{x}dx} \;=\;e^{3\ln x} \;=\;e^{\ln x^3} \;=\;x^3$

Got it?

$2)\;\;(\cos^2\!x - y\cos x)\,dx - (1 + \sin x)\,dy \:= \:0$

We have: . $\underbrace{(\cos^2\!x - y\cos x)}_M\,dx + \underbrace{(\text{-}1-\sin x)}_N\,dy \;=\;0$

. . . . . . . . $\frac{\partial M}{\partial y} \:=\:-\cos x \qquad\quad\frac{\partial N}{\partial x} \:=\:-\cos x$

The equation is exact!

5. Originally Posted by ThePerfectHacker
3)Divide by $x^2$ to bring to form as in #1.
I try thtat , and the equation become Bernoulli Eq. and I should use Substitution to make it become Linear D.E ?

Originally Posted by mr fantastic
4) is a second order DE with constant coefficients. Get the homogeneous solution and use the method of undetermined coefficients to construct a particular solution.
After I get the homogeneous solution then wath should I do, I mean, how to use the undetermined coefficient method? should I convert the "sinx " in to "e^ia thing"

And for number 5), the variation of parameters methode is the methode taht using wronskian ? Sorry i can't remember the name of the methods...

Thanks Again

6. Originally Posted by Singular
$
\begin{gathered}
3).x^2 \frac{{dy}}
{{dx}} + xy = xy^3 \hfill \\
\end{gathered}
$
Originally Posted by Singular
I try thtat , and the equation become Bernoulli Eq. and I should use Substitution to make it become Linear D.E ?
If you look at it again, you'll see that it's separable rather than Bernoulli.

$x^2y' + xy = xy^3$

$\frac{x^2y'}{x^2} + \frac{xy}{x^2} = \frac{xy^3}{x^2}$

$y' + \frac{y}{x} = \frac{y^3}{x}$

$y' = \frac{y^3}{x} - \frac{y}{x}$

$y' = \frac{y^3-y}{x}$

$\frac{dy}{y^3-y} = \frac{dx}{x}$.....

7. After I get the homogeneous solution then wath should I do, I mean, how to use the undetermined coefficient method? should I convert the "sinx " in to "e^ia thing
First I suggest you read the following article:

Method of undetermined coefficients - Wikipedia, the free encyclopedia

after reading this article you'll come to the conclusion that you should assume a particular solution of following form:

$
Ae^x + B\sin x + C\cos x
$

just plug it in the ode and equate the appropriate coefficients.

8. Originally Posted by Singular
[snip]
And for number 5), the variation of parameters methode is the methode taht using wronskian ?
[snip]
Yes.