User:Egm6321.f10.team3.Sudheesh/Homeworks/Homework2

= Problem 1 =

Solution
= Problem 2 =

Solution
= Problem 3 =

Solution
= Problem 4 - Create Exact Nonlinear First Order Ordinary Differential Equation (N1-ODE) =

Given
Conditions of Exactness:


 * For function $$F\left( . \right)$$ to be exact:


 * $$\left( 1 \right)$$ $$F\left( . \right)$$ must be in the form:


 * {| style="width:100%" border="0"

$$
 * $$F\left( x,y,{{y}^{1}} \right)=M\left( x,y \right)+N\left( x,y \right){{y}^{'}}$$
 * $$\displaystyle (Eq. 4.1)
 * $$\displaystyle (Eq. 4.1)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"


 * Where $$M\left( x,y \right)$$, and $$N\left( x,y \right)$$, are functions of $$x$$, and $$y$$, $$=\frac{dy}{dx}$$
 * }
 * }
 * }


 * $$\left( 2 \right)$$ $$M\left( x,y \right)$$, and $$N\left( x,y \right)$$ must satisfy:


 * {| style="width:100%" border="0"

$$
 * $$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$
 * $$\displaystyle (Eq. 4.2)
 * $$\displaystyle (Eq. 4.2)
 * }
 * }

Given function:


 * {| style="width:100%" border="0"

$$
 * $$\phi \left( x,y \right)={{x}^{2}}{{y}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}+\log \left( {{x}^{3}}{{y}^{2}} \right)=k$$
 * $$\displaystyle (Eq. 4.3)
 * $$\displaystyle (Eq. 4.3)
 * }
 * }

Find

 * $$\left( 1 \right)$$ Find function $$F\left( . \right)$$ such that $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}$$


 * $$\left( 2 \right)$$ Verify that $$F\left( x,y,{{y}^{1}} \right)$$ is an Exact N1-ODE


 * $$\left( 3 \right)$$ Invent three more Exact N1-ODE

Solution

 * $$\left( 1 \right)$$ Find function $$F\left( x,y,{{y}^{1}} \right)$$:


 * Derivative of $$\phi \left( x,y \right)$$ is given by
 * {| style="width:100%" border="0"

$$
 * $$\frac{d\phi }{dx}=\frac{\partial \phi }{\partial x}+\frac{\partial \phi }{\partial y}\cdot {{y}^{'}}$$
 * $$\displaystyle (Eq. 4.4)
 * $$\displaystyle (Eq. 4.4)
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\frac{\partial \phi }{\partial x}=\left[ 2x{{y}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}+\frac{1}\cdot 3{{x}^{2}}{{y}^{2}} \right]=2x{{y}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}+\frac{3}{x}$$
 * $$\displaystyle (Eq. 4.5)
 * $$\displaystyle (Eq. 4.5)
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\frac{\partial \phi }{\partial y}=\left[ \frac{3}{2}{{x}^{2}}{{y}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+\frac{1}\cdot 2{{x}^{3}}y \right]=\frac{3}{2}{{x}^{2}}{{y}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+\frac{2}{y}$$
 * $$\displaystyle (Eq. 4.6)
 * $$\displaystyle (Eq. 4.6)
 * }
 * Substituting Eq.4.5 & 4.6 into Eq.4.4 yields:
 * Substituting Eq.4.5 & 4.6 into Eq.4.4 yields:


 * {| style="width:100%" border="0"

$$
 * $$\frac{d\phi }{dx}=\left[ 2x{{y}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}+\frac{3}{x} \right]+\left[ \frac{3}{2}{{x}^{2}}{{y}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+\frac{2}{y} \right]\cdot {{y}^{'}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.7)
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.7)
 * }
 * Therefore:
 * Therefore:


 * {| style="width:100%" border="0"

$$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}=\underbrace{\left[ 2x{{y}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}+\frac{3}{x} \right]}_{M\left( x,y \right)}+\underbrace{\left[ \frac{3}{2}{{x}^{2}}{{y}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+\frac{2}{y} \right]}_{N\left( x,y \right)}\cdot {{y}^{'}}$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.8)
 * }
 * }


 * $$\left( 2 \right)$$ Verify that $$F\left( x,y,{{y}^{1}} \right)$$ is an Exact N1-ODE:


 * The function $$F\left( x,y,{{y}^{1}} \right)$$ is a first order nonlinear differential equation as its highest order is 1 and does not satisfy the condtion of superposition (linearity).
 * The function $$F\left( x,y,{{y}^{1}} \right)$$ given in Eq. 4.8 has the structure of Eq.4.1 and hence first condition of Exactness is satisfied.


 * {| style="width:100%" border="0"

$$
 * $$\frac{\partial M}{\partial y}=\frac{\partial }{\partial y}\left( 2x{{y}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}+\frac{3}{x} \right)=2x\cdot \frac{3}{2}{{y}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+0=3x{{y}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.9)
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.9)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$
 * $$\frac{\partial N}{\partial x}=\frac{\partial }{\partial x}\left( \frac{3}{2}{{x}^{2}}{{y}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+\frac{2}{y} \right)=\frac{3}{2}\cdot 2x{{y}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+0=3x{{y}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.10)
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.10)
 * }
 * Eq.4.9 and 4.10 implies,
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$
 * $$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.11)
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.11)
 * }
 * Eq.4.11 satisfy the second condition of Exactness.
 * Eq.4.11 satisfy the second condition of Exactness.


 * Since both conditions of exactness are satisifed,
 * {| style="width:100%" border="0"


 * $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}=\underbrace{\left[ 2x{{y}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}+\frac{3}{x} \right]}_{M\left( x,y \right)}+\underbrace{\left[ \frac{3}{2}{{x}^{2}}{{y}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+\frac{2}{y} \right]}_{N\left( x,y \right)}\cdot {{y}^{'}}$$ is an Exact N1-ODE
 * }
 * $$\left( 3 \right)$$ Invent three more Exact N1-ODE:
 * Example 1:
 * Consider, $$\phi \left( x,y \right)=3x\cdot {{e}^{y}}+4{{y}^{2}}\cdot {{e}^{x}}= k $$
 * Example 1:
 * Consider, $$\phi \left( x,y \right)=3x\cdot {{e}^{y}}+4{{y}^{2}}\cdot {{e}^{x}}= k $$


 * {| style="width:100%" border="0"


 * $$\frac{\partial \phi }{\partial x}=3\cdot {{e}^{y}}+4{{y}^{2}}\cdot {{e}^{x}}\Rightarrow M(x,y)$$
 * }
 * {| style="width:100%" border="0"
 * }
 * {| style="width:100%" border="0"


 * $$\frac{\partial \phi }{\partial y}=3x\cdot {{e}^{y}}+8y\cdot {{e}^{x}}\Rightarrow N\left( x,y \right)$$
 * }
 * Now consider a function $$F\left( x,y,{{y}^{1}} \right)$$ such that $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}$$
 * }
 * Now consider a function $$F\left( x,y,{{y}^{1}} \right)$$ such that $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}$$


 * {| style="width:100%" border="0"


 * $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}=M\left( x,y \right)+N\left( x,y \right)\cdot {{y}^{'}}=0$$
 * }
 * {| style="width:100%" border="0"
 * }
 * {| style="width:100%" border="0"

$$F\left( x,y,{{y}^{1}} \right)=\left[ 3\cdot {{e}^{y}}+4{{y}^{2}}\cdot {{e}^{x}} \right]+\left[ 3x\cdot {{e}^{y}}+8y\cdot {{e}^{x}} \right]\cdot {{y}^{'}}=0$$ $$
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.12)
 * }
 * This is an Exact N1-ODE
 * This is an Exact N1-ODE


 * Example 2:
 * Consider, $$\phi \left( x,y \right)=5y\log \left( {{x}^{2}} \right)+3{{x}^{2}}\sin \left( {{y}^{3}} \right)=k$$


 * {| style="width:100%" border="0"


 * $${{\phi }_{x}}\left( x,y \right)=5y\cdot \left( \frac{1}\cdot 2x \right)+6x\sin \left( {{y}^{3}} \right)=10\frac{y}{x}+6x\sin \left( {{y}^{3}} \right)\Rightarrow M(x,y)$$
 * }
 * {| style="width:100%" border="0"
 * }
 * {| style="width:100%" border="0"


 * $${{\phi }_{y}}\left( x,y \right)=5\log \left( {{x}^{2}} \right)+3{{x}^{2}}\left( \cos \left( {{y}^{3}} \right)\cdot 3{{y}^{2}} \right)=5\log \left( {{x}^{2}} \right)+9{{x}^{2}}{{y}^{2}}\cos \left( {{y}^{3}} \right)\Rightarrow N\left( x,y \right)$$
 * }
 * Now consider a function $$F\left( x,y,{{y}^{1}} \right)$$ such that $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}$$
 * }
 * Now consider a function $$F\left( x,y,{{y}^{1}} \right)$$ such that $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}$$


 * {| style="width:100%" border="0"


 * $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}=M\left( x,y \right)+N\left( x,y \right)\cdot {{y}^{'}}=0$$
 * }
 * {| style="width:100%" border="0"
 * }
 * {| style="width:100%" border="0"

$$F\left( x,y,{{y}^{'}} \right)=\left[ 10\frac{y}{x}+6x\sin \left( {{y}^{3}} \right) \right]+\left[ 5\log \left( {{x}^{2}} \right)+9{{x}^{2}}{{y}^{2}}\cos \left( {{y}^{3}} \right) \right]{{y}^{'}}=0$$ $$
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.13)
 * }
 * This is an Exact N1-ODE
 * This is an Exact N1-ODE


 * Example 3:
 * Consider, $$\phi \left( x,y \right)=7{{x}^{3}}\cdot \cos \left( {{y}^{3}} \right)+2{{y}^{2}}{{e}^{{{x}^{2}}\cdot \sin y}}=k$$


 * {| style="width:100%" border="0"

& {{\phi }_{x}}\left( x,y \right)=7\times 3{{x}^{2}}\cdot \cos \left( {{y}^{3}} \right)+2{{y}^{2}}\cdot {{e}^{{{x}^{2}}\cdot \sin y}}\cdot \left( 2x\cdot \sin y \right)=21{{x}^{2}}\cdot \cos \left( {{y}^{3}} \right)+4x{{y}^{2}}\cdot \sin y\cdot {{e}^{{{x}^{2}}\cdot \sin y}}\Rightarrow M(x,y) \\ & \\ \end{align}$$
 * $$\begin{align}
 * $$\begin{align}


 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"


 * $${{\phi }_{y}}\left( x,y \right)=7{{x}^{3}}\cdot \left[ -\sin \left( {{y}^{3}} \right)\cdot 3{{y}^{2}} \right]+2\left[ 2y\cdot {{e}^{{{x}^{2}}\cdot \sin y}}+{{y}^{2}}\cdot {{e}^{{{x}^{2}}\cdot \sin y}}\cdot {{x}^{2}}\cdot \cos y \right]=-21{{x}^{3}}{{y}^{2}}\sin \left( {{y}^{3}} \right)+4y{{e}^{{{x}^{2}}\cdot \sin y}}+2{{x}^{2}}{{y}^{2}}\cos y\cdot {{e}^{{{x}^{2}}\cdot \sin y}}\Rightarrow N\left( x,y \right)$$
 * }
 * Now consider a function $$F\left( x,y,{{y}^{1}} \right)$$ such that $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}$$
 * }
 * Now consider a function $$F\left( x,y,{{y}^{1}} \right)$$ such that $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}$$


 * {| style="width:100%" border="0"


 * $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}=M\left( x,y \right)+N\left( x,y \right)\cdot {{y}^{'}}=0$$
 * }
 * {| style="width:100%" border="0"
 * }
 * {| style="width:100%" border="0"

$$\left( x,y,{{y}^{'}} \right)=\left[ 21{{x}^{2}}\cdot \cos \left( {{y}^{3}} \right)+4x{{y}^{2}}\cdot \sin y\cdot {{e}^{{{x}^{2}}\cdot \sin y}} \right]+\left[ -21{{x}^{3}}{{y}^{2}}\sin \left( {{y}^{3}} \right)+4y{{e}^{{{x}^{2}}\cdot \sin y}}+2{{x}^{2}}{{y}^{2}}\cos y\cdot {{e}^{{{x}^{2}}\cdot \sin y}} \right]{{y}^{'}}=0$$ $$
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.14)
 * }
 * This is an Exact N1-ODE
 * This is an Exact N1-ODE