User:Jiangpx/6321 hw2

=prob1 - 7-1=

=prob 2 - 7-1: verify (7) is a N1-ODE=

Given
$$\begin{align} & M(x,y)=4{{x}^{7}}+\sin y \\ & N(x,y)={{x}^{2}}{{y}^{3}} \\ \end{align}$$

Find
$$(4{{x}^{7}}+\sin y)+({{x}^{2}}{{y}^{3}})y'=0$$ is a N1-ODE

Solution
In general, N1-ODE has the form of $$M(x,y)+N(x,y)y'=0$$,

in which, $$y'$$ tells that it is of order 1 and the coefficients M and N are nonlinear - they violate the definition of linear operator, defined as follows:


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

$$\displaystyle L(cx)=cL(x)$$
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


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

$$\displaystyle L(x+y)=L(x)+L(y)$$
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Hence, the given equation is a N1-ODE.

=prob 3 - 8-1=

=prob 4 - 8-2=

=prob 5 - 9-2=

=prob 6 - 9-3=

Given
$$y(x)={{\sin }^{-1}}(k-15{{x}^{5}})$$

$$\begin{align} & M(x,y)=75{{x}^{4}} \\ & N(x,y)=\cos y \\ \end{align}$$

Verify
$$y(x)={{\sin }^{-1}}(k-15{{x}^{5}})$$ satisfies $$M(x,y)+N(x,y)y'=0$$

Solution
The ODE to be solved is

$$75{{x}^{4}}+(\cos y)y'=0$$

and from $$y(x)={{\sin }^{-1}}(k-15{{x}^{5}})$$

we get $$\sin y=k-15{{x}^{5}}$$

Differentiate both sides, $$\cos y\cdot y'=-75{{x}^{4}}$$, which is the ODE to be solved.

Hence, verified.

=prob 7 - 10-3 & 11-1= =prob 8 - 11-2= =prob 9 - 12-2=