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Math Help - Ode

  1. #1
    Junior Member Singular's Avatar
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    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 ....


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

    Thank you very much
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  2. #2
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    Quote Originally Posted by Singular View Post

    <br />
\begin{gathered}<br />
  1).\frac{{dy}}<br />
{{dx}} + 3\frac{y}<br />
{x} = 6x^2  \hfill \\<br />
  2).((\cos ^2 x - y\cos x))dx - (1 + \sin x)dy = 0 \hfill \\<br />
  3).x^2 \frac{{dy}}<br />
{{dx}} + xy = xy^3  \hfill \\<br />
  4).\frac{{d^2 y}}<br />
{{dx^2 }} - 2\frac{{dy}}<br />
{{dx^{} }} - 3y = 2e^x  - 10\sin x \hfill \\<br />
  5).\frac{{d^2 y}}<br />
{{dx^2 }} + y = \tan x\sec x \hfill \\ <br />
\end{gathered}<br /> <br />
    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.
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  3. #3
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    Quote Originally Posted by Singular View Post
    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 ....


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

    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.
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  4. #4
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    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!

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  5. #5
    Junior Member Singular's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    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 ?

    Quote Originally Posted by mr fantastic View Post
    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
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  6. #6
    Super Member wingless's Avatar
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    Quote Originally Posted by Singular View Post
    <br />
\begin{gathered}<br />
  3).x^2 \frac{{dy}}<br />
{{dx}} + xy = xy^3  \hfill \\<br />
\end{gathered}<br />
    Quote Originally Posted by Singular View Post
    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}.....
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  7. #7
    Senior Member Peritus's Avatar
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    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:

    <br />
Ae^x  + B\sin x + C\cos x<br />

    just plug it in the ode and equate the appropriate coefficients.
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  8. #8
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    Quote Originally Posted by Singular View Post
    [snip]
    And for number 5), the variation of parameters methode is the methode taht using wronskian ?
    [snip]
    Yes.
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