User:Egm6321.f10.team03.sigillo/Hwk2

= Problem 6=

Given
Suppose a function $$\phi $$ can be found to be a reduced order solution and its derivative is equal to:
 * {| style="width:100%" border="0"

$$\frac{d\phi (x,y,{y}')}{dx}=F(x,y,{y}')=75{{x}^{4}}+\cos \left( y \right){y}'=0$$ $$ and the explicit solution for $$\,y(x)$$ is given as:
 * $$\,(Eq. 6.1)
 * style= |
 * }
 * {| style="width:100%" border="0"

$$y(x)={{\sin }^{-1}}\left( k-15{{x}^{5}} \right)$$. $$
 * $$\,(Eq. 6.2)
 * style= |
 * }

Find
It has to be proven that (Eq. 6.2) satisfies (Eq. 6.1)

Solution
In general $$F(x,y,{y}')$$ can be written as:
 * {| style="width:100%" border="0"

$$\,F(x,y,{y}')=M(x,y)+N(x,y){y}'=0$$ $$ where from (Eq. 6.1) it is known that
 * $$\,(Eq. 6.3)
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,M(x,y)=75{{x}^{4}}$$ $$ and
 * $$\,(Eq. 6.4)
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,N(x,y)=\cos \left( y \right)$$ $$ The chosen approach is to derivate $$y(x)$$ ,insert the result into (Eq. 6.2) and check if it satisfies the condition $$F(x,y,{y}')=0$$.In order to have a better overview during deferentiating, let's define
 * $$\,(Eq. 6.5)
 * style= |
 * }
 * {| style="width:100%" border="0"

$$u(x)=\left( k-15{{x}^{5}} \right)$$ $$ The derivative is defined to be:
 * $$\,(Eq. 6.6)
 * style= |
 * }
 * {| style="width:100%" border="0"

$${y}'=\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$$ $$ which in the particular case it is:
 * $$\,(Eq. 6.7)
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\frac{dy}{du}=\frac{1}{\sqrt{1-{{u}^{2}}}}$$ $$ and
 * $$\,(Eq. 6.8)
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\frac{du}{dx}=-75{{x}^{4}}$$ $$ Inserting (Eq. 6.8) and (Eq. 6.9) into (Eq. 6.7) while substituiting $$u$$ back, (Eq. 6.7) can be rewritten as:
 * $$\,(Eq. 6.9)
 * style= |
 * }
 * {| style="width:100%" border="0"

$${y}'=\frac{-75{{x}^{4}}}{\sqrt{1-{{\left( k-15{{x}^{5}} \right)}^{2}}}}$$ $$ The key-point of this Problem in terms of quickness, lays in the fact that is has to be recognized that $${{\left( k-15{{x}^{5}} \right)}^{2}}$$ is the same like $${{sin }^{2}}\left( y \right)$$, hence (Eq. 6.10) becomes:
 * $$\,(Eq. 6.10)
 * style= |
 * }
 * {| style="width:100%" border="0"

$${y}'=\frac{-75{{x}^{4}}}{\sqrt{1-{{\sin }^{2}}\left( y \right)}}=\frac{-75{{x}^{4}}}{\sqrt{{{\cos }^{2}}\left( y \right)}}=\frac{-75{{x}^{4}}}{\cos \left( y \right)}$$ $$ Now substituting y' in (Eq. 6.1) with (Eq. 6.11) we get:
 * $$\,(Eq. 6.11)
 * style= |
 * }
 * {| style="width:100%" border="0"

$$75{{x}^{4}}+\cos \left( y \right)\frac{-75{{x}^{4}}}{\cos \left( y \right)}=75{{x}^{4}}-75{{x}^{4}}=0$$ $$ :{| style="width:100%" border="0" Left-Hand-Side and Right-Hand-Side of the equation are the same, hence (Eq. 6.2) satisfies (Eq. 6.1) !
 * $$\,(Eq. 6.12)
 * style= |
 * }
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * }
 * }

= Problem 5=

Given
The general form of a N1_ODE is given as being:
 * {| style="width:100%" border="0"

$$F(x,y,{y}')=M(x,y)+N(x,y)f\left( {{y}'} \right)=0$$
 * $$\,(Eq. 5.1) $$
 * style= |
 * }

Find
To be found is a function $$f\left( {{y}'} \right)$$ such that there is no analytical solution to $$f\left( {{y}'} \right)=\frac{-M}{N}$$ , hence a non exact N1_ODE has to be found.

Solution
Let's set the equation to be analyzed as:
 * {| style="width:100%" border="0"

$$2x{{y}^{2}}+\cos \left( 3y \right)x\cdot {y}'\exp \left( {{y}'} \right)=0$$ where the three terms of the general equation (Eq. 5.1) in this application caser can be found as:
 * $$\,(Eq. 5.2) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$M\left( x,y \right)=2x{{y}^{2}}\quad \quad N\left( x,y \right)=\cos \left( 3y \right)x\quad \quad f({y}')={y}'\cdot \exp \left( {{y}'} \right)$$ Since the first conditoin for an exact N1_ODE is defined to be:
 * style= |
 * }
 * }
 * {| style="width:100%" border="0"

$$\,M(x,y)+N(x,y){y}'=0$$ It's needed to put (Eq. 5.2) in the form of (Eq. 5.3). Thus the next step is to solve (Eq. 5.2) for y' :
 * $$\,(Eq. 5.3) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${y}'\cdot \exp \left( {{y}'} \right)=\frac{-2x{{y}^{2}}}{\cos \left( 3y \right)x}=\frac{-2{{y}^{2}}}{\cos \left( 3y \right)}$$ trying to get y' out of the exp-function we apply the logarithm-funtion:
 * $$\,(Eq. 5.4) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\ln \left( {y}'\cdot \exp \left[ {{y}'} \right] \right)=\ln \left( \frac{-2{{y}^{2}}}{\cos \left( 3y \right)} \right)=\ln \left( {{y}'} \right)+\ln \left( \exp \left[ {{y}'} \right] \right)$$
 * <p style="text-align:right;">$$\,(Eq. 5.5) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\ln \left( {{y}'} \right)+{y}'=\ln \left( \frac{-2{{y}^{2}}}{\cos \left( 3y \right)} \right)$$ :{| style="width:90%" border="0" As it can be seen, there is on analytical solution for y' in (Eq. 5.6), this means that (Eq. 5.2) can not be put in the form of (Eq. 5.3), hence it is NOT an exact N1_ODE !
 * <p style="text-align:right;">$$\,(Eq. 5.6) $$
 * style= |
 * }
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * }
 * }

= Problem 1 - Non-linearity in general of N1_ODE =

Given
Given is the general form of a N1_ODE:
 * {| style="width:100%" border="0"

$$M(x,y)+N(x,y)\frac{dy}{dx}=0$$
 * <p style="text-align:right;">$$\,(Eq. 1.1) $$
 * style= |
 * }

Find
It has to be shown, that IN GENERAL (Eq. 1.1) is non-linear

Solution
To show that (Eq. 1.1) is in general non-linear it needs to meet the non-linearity condition, which is:
 * {| style="width:100%" border="0"

$$F\left( \alpha u+\beta v \right)\ne F\left( \alpha u \right)+F\left( \beta v \right)=\alpha F\left( u \right)+\beta F\left( v \right)$$ Applying the theoreme above to (Eq. 1.1), we have to see if the following equation is true:
 * <p style="text-align:right;">$$\,(Eq. 1.2) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$cM(x,y)+cN(x,y)\frac{dy}{dx}\overset{?}{\mathop{=}}\,M(x,cy)+N(x,cy)\frac{d(cy)}{dx}$$ Now, since N(x,y) can be ANY function, like a cos(2xy) which is non-linear, it turns out that (Eq. 1.3) is not true and is IN GENERAL not linear: :{| style="width:30%" border="0" $$M(x,y)+N(x,y)\frac{dy}{dx}=0\ is\ in\ general\ non\ linear!$$ For particular solutions it might be the case, that (Eq. 1.1) is linear, but that depends on the form of the functions M(x,y) and N(x,y).
 * <p style="text-align:right;">$$\,(Eq. 1.3) $$
 * style= |
 * }
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * }
 * }