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Math Help - Hard differential equations

  1. #1
    Newbie LaraSoft's Avatar
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    Unhappy Hard differential equations

    Hello, boys .
    In my home work is to solve these 3 equations , and a textual problem , but the second day I can not solve them. My professor said that these tasks are very easy, I just need to better analyze, but I think he has a good sense of humor.

    1. Find a nontrivial solution of differential equation 2yy'' + {\left( {y'} \right)^2} + {\left( {y'} \right)^4} = 0.

    2. Solve the differential equation \left( {3y - {x^3}} \right)y' = 2x + 3{x^2}y.

    3. Using Laplace transform, solve the differential equation y'' + 2y' + 2y = \left| {\sin t} \right| if y\left( 0 \right) = 0 and y'\left( 0 \right) = 0.

    Please help me.
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  2. #2
    MHF Contributor
    Jester's Avatar
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    Quote Originally Posted by LaraSoft View Post
    Hello, boys .
    In my home work is to solve these 3 equations , and a textual problem , but the second day I can not solve them. My professor said that these tasks are very easy, I just need to better analyze, but I think he has a good sense of humor.

    1. Find a nontrivial solution of differential equation 2yy'' + {\left( {y'} \right)^2} + {\left( {y'} \right)^4} = 0.

    2. Solve the differential equation \left( {3y - {x^3}} \right)y' = 2x + 3{x^2}y.

    3. Using Laplace transform, solve the differential equation y'' + 2y' + 2y = \left| {\sin t} \right| if y\left( 0 \right) = 0 and y'\left( 0 \right) = 0.

    Please help me.
    Here's some help to get you going.

    On the first ODE, there's no x so if you let

    y' = p then y'' = \frac{dp}{dx} = \frac{dp}{dy} \frac{dy}{dx} = p \frac{dp}{dy}

    so

    2 \,y \,p \,\frac{dp}{dy} + p^2 + p^4 = 0 separable.

    On the second, if you re-write it as

    \left( 2x +3x^2y \right) \,dx + \left( x^3 - 3y \right) \, dy = 0

    you'll find it's exact.
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by LaraSoft View Post
    Hello, boys .
    In my home work is to solve these 3 equations , and a textual problem , but the second day I can not solve them. My professor said that these tasks are very easy, I just need to better analyze, but I think he has a good sense of humor.

    1. Find a nontrivial solution of differential equation 2yy'' + {\left( {y'} \right)^2} + {\left( {y'} \right)^4} = 0.

    2. Solve the differential equation \left( {3y - {x^3}} \right)y' = 2x + 3{x^2}y.

    3. Using Laplace transform, solve the differential equation y'' + 2y' + 2y = \left| {\sin t} \right| if y\left( 0 \right) = 0 and y'\left( 0 \right) = 0.

    Please help me.
    2. Solve the differential equation \left( {3y - {x^3}} \right)y' = 2x + 3{x^2}y.

    \left( {3y - {x^3}} \right)y' = 2x + 3{x^2}y \Leftrightarrow \left( {2x + 3{x^2}y} \right)dx + \left( {{x^3} - 3y} \right)<br />
dy = 0 \Leftrightarrow P\left( {x,y} \right)dx + Q\left( {x,y} \right)dy = 0

    So {P_y}^\prime  = {Q_x}^\prime hence the expression is a full differential of some implicit function F\left( {x,y} \right) = C.

    F\left( {x,y} \right) = \int\limits_{\left( {{x_0},{y_0}} \right)}^{\left( {x,y} \right)} {P\left( {\xi ,\eta } \right)d\xi }  + Q\left( {\xi ,\eta } \right)d\eta  = \int\limits_{{x_0}}^x {P\left( {\xi ,{y_0}} \right)d\xi }  + \int\limits_{{y_0}}^y {Q\left( {x,\eta } \right)d\eta }  =

    = \int\limits_{{x_0}}^x {\left( {2\xi  + 3{\xi ^2}{y_0}} \right)d\xi }  + \int\limits_{{y_0}}^y {\left( {{x^3} - 3\eta } \right)d\eta }  = \left. {\left( {{\xi ^2} + {\xi ^3}{y_0}} \right)} \right|_{{x_0}}^x + \left. {\left( {{x^3}\eta  - \frac{3}{2}{\eta ^2}} \right)} \right|_{{y_0}}^y =

    = \left( {{x^2} + {x^3}{y_0} - x_0^2 - x_0^3{y_0}} \right) + \left( {{x^3}y - \frac{3}{2}{y^2} - {x^3}y_0^2 + \frac{3}{2}y_0^2} \right) =

    = \left( {{x^2} + {x^3}y - \frac{3}{2}{y^2}} \right) - \left( {x_0^2 + x_0^3{y_0} - \frac{3}{2}y_0^2} \right) = C.

    F\left( {x,y} \right) = \boxed{{x^2} + {x^3}y - \frac{3}{2}{y^2} = C}
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  4. #4
    Senior Member DeMath's Avatar
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    3. Using Laplace transform, solve the differential equation y'' + 2y' + 2y = \left| {\sin t} \right| if y\left( 0 \right) = 0 and y'\left( 0 \right) = 0.
    y\left( t \right)\underset{\raise0.3em\hbox{$\smash{\scripts  criptstyle\cdot}$}}{\dot  = } \overline {y\left( s \right)}

    y'\left( t \right)\underset{\raise0.3em\hbox{$\smash{\scripts  criptstyle\cdot}$}}{\dot  = } s\overline {y\left( s \right)}  - y\left( 0 \right) = s\overline {y\left( s \right)}

    y''\left( t \right)\underset{\raise0.3em\hbox{$\smash{\scripts  criptstyle\cdot}$}}{\dot  = } {s^2}\overline {y\left( s \right)}  - y\left( 0 \right)s - y'\left( 0 \right) = {s^2}\overline {y\left( s \right)}

    \left| {\sin t} \right|\underset{\raise0.3em\hbox{$\smash{\scripts  criptstyle\cdot}$}}{\dot  = } \frac{{1 + {e^{ - \pi s}}}}{{1 - {e^{ - \pi s}}}} \cdot \frac{1}{{1 + {s^2}}}

    So, we have

    {s^2}\overline {y\left( s \right)}  + 2s\overline {y\left( s \right)}  + 2\overline {y\left( s \right)}  = \frac{{1 + {e^{ - \pi s}}}}{{1 - {e^{ - \pi s}}}} \cdot \frac{1}{{1 + {s^2}}} \Leftrightarrow

    \Leftrightarrow \left( {{s^2} + 2s + 2} \right)\overline {y\left( s \right)}  = \frac{{1 + {e^{ - \pi s}}}}{{1 - {e^{ - \pi s}}}} \cdot \frac{1}{{1 + {s^2}}} \Leftrightarrow

    \Leftrightarrow \overline {y\left( s \right)}  = \frac{1}{{\left( {1 + {s^2}} \right)\left[ {{{\left( {s + 1} \right)}^2} + 1} \right]}} \cdot \frac{{1 + {e^{ - \pi s}}}}{{1 - {e^{ - \pi s}}}}.

    Next we will use the third decomposition theorem. In this case we can use this theorem, because \mathop {\lim }\limits_{s \to \infty } \overline {y\left( s \right)}  = 0 and \mathop {\lim }\limits_{s \to \infty } \overline {y\left( s \right)}  = 0 uniformly on the entire plane except for poles.

    The function \overline {y\left( s \right)} has poles s =  \pm i,{\text{ }}s =  - 1 \pm i,{\text{ }}s = 2ni{\text{ }}\left( {n \in \mathbb{Z}} \right).

    Next we will find residues of the function {e^{st}}\overline {y\left( s \right)} in these poles.

    1.{\text{ }}\mathop {{\text{Res}}}\limits_{s = i} \left\{ {{e^{st}}\overline {y\left( s \right)} } \right\} = \mathop {\lim }\limits_{s \to i} \left[ {\frac{{s - 1}}{{{s^2} + 1}} \cdot \frac{{{e^{st}}}}{{{{\left( {s + 1} \right)}^2} + 1}} \cdot \frac{{1 + {e^{ - \pi s}}}}{{1 - {e^{ - \pi s}}}}} \right] = 0.

    Similarly, \mathop {{\text{Res}}}\limits_{s =  - i} \left\{ {{e^{st}}\overline {y\left( s \right)} } \right\} = 0.

    2.{\text{ }}\mathop {{\text{Res}}}\limits_{s =  - 1 + i} \left\{ {{e^{st}}\overline {y\left( s \right)} } \right\} = \mathop {\lim }\limits_{s \to  - 1 + i} \left[ {\frac{1}{{{s^2} + 1}} \cdot \frac{{\left( {s + 1 - i} \right){e^{st}}}}{{{{\left( {s + 1} \right)}^2} + 1}} \cdot \frac{{1 + {e^{ - \pi s}}}}{{1 - {e^{ - \pi s}}}}} \right] =  - \tanh \left( {\frac{\pi }{2}} \right)\frac{{2 - i}}{{10}}{e^{ - t}}{e^{it}}.

    \mathop {{\text{Res}}}\limits_{s =  - 1 - i} \left\{ {{e^{st}}\overline {y\left( s \right)} } \right\} =  - \tanh \left( {\frac{\pi }{2}} \right)\frac{{2 + i}}{{10}}{e^{ - t}}{e^{ - it}}.

    Sum these of two residues

    \mathop {{\text{Res}}}\limits_{s =  - 1 + i} \left\{ {{e^{st}}\overline {y\left( s \right)} } \right\} + \mathop {{\text{Res}}}\limits_{s =  - 1 - i} \left\{ {{e^{st}}\overline {y\left( s \right)} } \right\} =  - \frac{1}{5}\tanh \left( {\frac{\pi }{2}} \right)\left( {2\cos t + \sin t} \right){e^{ - t}}.

    3. We will find a residue in the point s = 2ni

    if n=0

    \mathop {{\text{Res}}}\limits_{s = 0} \left\{ {{e^{st}}\overline {y\left( s \right)} } \right\} = \mathop {\lim }\limits_{s \to 0} \left[ {\frac{1}{{{s^2} + 1}} \cdot \frac{{{e^{st}}}}{{{{\left( {s + 1} \right)}^2} + 1}} \cdot \left( {1 + {e^{ - \pi s}}} \right) \cdot \frac{s}{{1 - {e^{ - \pi s}}}}} \right] = \mathop {\lim }\limits_{s \to 0} \frac{s}{{1 - {e^{ - \pi s}}}} = \frac{1}{\pi }.

    if n > 0

    \mathop {{\text{Res}}}\limits_{s = 2ni} \left\{ {{e^{st}}\overline {y\left( s \right)} } \right\} = \mathop {\lim }\limits_{s \to 2ni} \left[ {\frac{1}{{{s^2} + 1}} \cdot \frac{{{e^{st}}}}{{{{\left( {s + 1} \right)}^2} + 1}} \cdot \left( {1 + {e^{ - \pi s}}} \right) \cdot \frac{{s - 2ni}}{{1 - {e^{ - \pi s}}}}} \right] = \frac{{\left( {2{n^2} - 1} \right) + 2ni}}{{\left( {4{n^2} - 1} \right)\left( {4{n^2} + 1} \right)}} \cdot \frac{{{e^{2nit}}}}{\pi }.

    if n < 0

    \mathop {{\text{Res}}}\limits_{s =  - 2ni} \left\{ {{e^{st}}\overline {y\left( s \right)} } \right\} = \frac{{\left( {2{n^2} - 1} \right) - 2ni}}{{\left( {4{n^2} - 1} \right)\left( {4{n^2} + 1} \right)}} \cdot \frac{{{e^{ - 2nit}}}}{\pi }.

    Sum of these two residues

    \mathop {{\text{Res}}}\limits_{s = 2ni} \left\{ {{e^{st}}\overline {y\left( s \right)} } \right\} + \mathop {{\text{Res}}}\limits_{s =  - 2ni} \left\{ {{e^{st}}\overline {y\left( s \right)} } \right\} = \frac{2}{\pi } \cdot \frac{{\left( {2{n^2} - 1} \right)\cos \left( {2nt} \right) - 2n\sin \left( {2nt} \right)}}{{\left( {4{n^2} - 1} \right)\left( {4{n^2} + 1} \right)}}.

    Finally, we have

    y\left( t \right) = \frac{1}{\pi } - \frac{{{e^{ - t}}}}{5}\tanh \left( {\frac{\pi }{2}} \right)\left( {2\cos t + \sin t} \right) + \frac{2}{\pi } \sum\limits_{n = 1}^\infty  {\frac{{\left( {2{n^2} - 1} \right)\cos \left( {2nt} \right) - 2n\sin \left( {2nt} \right)}}{{\left( {4{n^2} - 1} \right)\left( {4{n^2} + 1} \right)}}}.
    Last edited by DeMath; February 7th 2009 at 11:28 AM.
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