1. ## 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 $\displaystyle 2yy'' + {\left( {y'} \right)^2} + {\left( {y'} \right)^4} = 0$.

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

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

2. Originally Posted by LaraSoft
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 $\displaystyle 2yy'' + {\left( {y'} \right)^2} + {\left( {y'} \right)^4} = 0$.

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

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

Here's some help to get you going.

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

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

so

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

On the second, if you re-write it as

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

you'll find it's exact.

3. Originally Posted by LaraSoft
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 $\displaystyle 2yy'' + {\left( {y'} \right)^2} + {\left( {y'} \right)^4} = 0$.

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

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

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

$\displaystyle \left( {3y - {x^3}} \right)y' = 2x + 3{x^2}y \Leftrightarrow \left( {2x + 3{x^2}y} \right)dx + \left( {{x^3} - 3y} \right)$ $\displaystyle dy = 0 \Leftrightarrow P\left( {x,y} \right)dx + Q\left( {x,y} \right)dy = 0$

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

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

$\displaystyle = \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$$\displaystyle + \left. {\left( {{x^3}\eta - \frac{3}{2}{\eta ^2}} \right)} \right|_{{y_0}}^y = \displaystyle = \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) = \displaystyle = \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. \displaystyle F\left( {x,y} \right) = \boxed{{x^2} + {x^3}y - \frac{3}{2}{y^2} = C} 4. 3. Using Laplace transform, solve the differential equation \displaystyle y'' + 2y' + 2y = \left| {\sin t} \right| if \displaystyle y\left( 0 \right) = 0 and \displaystyle y'\left( 0 \right) = 0. \displaystyle y\left( t \right)\underset{\raise0.3em\hbox{\smash{\scripts criptstyle\cdot}}}{\dot = } \overline {y\left( s \right)} \displaystyle 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)} \displaystyle 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)} \displaystyle \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 \displaystyle {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 \displaystyle \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 \displaystyle \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 \displaystyle \mathop {\lim }\limits_{s \to \infty } \overline {y\left( s \right)} = 0 and \displaystyle \mathop {\lim }\limits_{s \to \infty } \overline {y\left( s \right)} = 0 uniformly on the entire plane except for poles. The function \displaystyle \overline {y\left( s \right)} has poles \displaystyle 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 \displaystyle {e^{st}}\overline {y\left( s \right)} in these poles. \displaystyle 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, \displaystyle \mathop {{\text{Res}}}\limits_{s = - i} \left\{ {{e^{st}}\overline {y\left( s \right)} } \right\} = 0. \displaystyle 2.{\text{ }}\mathop {{\text{Res}}}\limits_{s = - 1 + i} \left\{ {{e^{st}}\overline {y\left( s \right)} } \right\} =$$\displaystyle \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}}.$

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

Sum these of two residues

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

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

if $\displaystyle n=0$

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

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

Sum of these two residues

$\displaystyle \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\}$$\displaystyle = \frac{2}{\pi } \cdot \frac{{\left( {2{n^2} - 1} \right)\cos \left( {2nt} \right) - 2n\sin \left( {2nt} \right)}}{{\left( {4{n^2} - 1} \right)\left( {4{n^2} + 1} \right)}}. Finally, we have \displaystyle y\left( t \right) = \frac{1}{\pi } - \frac{{{e^{ - t}}}}{5}\tanh \left( {\frac{\pi }{2}} \right)\left( {2\cos t + \sin t} \right) + \frac{2}{\pi }$$\displaystyle \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)}}}.$