User:Egm6321.f11.team2.kim/PR2

= Problem 2.2- L2 ODE CC =

Given
$$\displaystyle y''+ y'y(x) = x $$ $$\displaystyle p(x):=y'(x)$$ $$\displaystyle p' + p = x$$(2.2.1) Solution is that $$\displaystyle p(x)= k_1 e^{-x}+x-1 $$(2.2.2)

Find
Verify that eq (2.2.2) is indeed solution for eq (2.2.1).

Solution
Equation 2.2.1 is that $$\displaystyle p' + p = x$$(2.2.1) And now Let $$\displaystyle T = x - p$$ $$\displaystyle \frac {dT}{dx}= 1 - p' $$ $$\displaystyle p'=1- \frac {dT}{dx}$$ Plug in eq (2.2.1) The equation 2.2.1 is following that $$\displaystyle 1- \frac{dT}{dx} = t$$ $$\displaystyle 1 - t = \frac {dT}{dx} $$ $$\displaystyle dx = -\frac {dT}{(t-1)}$$ $$\displaystyle -k - x = ln(t-1)$$ $$\displaystyle t - 1 = e^{-k-x}$$ $$ \displaystyle t= 1 - ke^{-x}$$ $$ \displaystyle p(x)= x - t = ke^{-x} + x - 1$$(2.2.2) It shows that eq (2.2.2) is indeed solution for eq (2.2.1).

= Problem 2.8 - First exactness condition =

Given
First exactness condition is that $$\displaystyle M(x,y) + N(x,y)\frac {dy}{dx} = 0$$

Find
Does the following N1-ODE satisfy the first exactness condition? $$\displaystyle M(x,y)cosy' + N(x,y)logy' = 0$$(2.8.1)

Solution
In order to satisfy the first exactness condition, it must be in the form of: $$\displaystyle M(x,y) + N(x,y)\frac {dy}{dx} = 0$$ $$\displaystyle y'= -\frac{M(x,y)}{N(x,y)}$$(2.8.2) So rearrange eq (2.8.1) to fit eq (2.8.2) $$\displaystyle M(x,y)cosy' + N(x,y)logy' = 0$$ $$\displaystyle M(x,y) + N(x,y)\frac{logy'}{cosy'}=0$$ $$\displaystyle f(y')=\frac{logy'}{cosy'}$$ $$\displaystyle f(y')$$ is any function of y'. If the function $$\displaystyle f(y')$$ has no explicit inverse $$\displaystyle f^{-1}(y'),$$ then it can not satisfy in the form of eq (2.8.2) In this case, $$\displaystyle f(y')$$ does not have exact inverse $$\displaystyle f^{-1}(y')$$ Thus, This does not satisfy the first exactness condition.

= Problem 2.14 - General L1-ODE-VC =

Given
$$\displaystyle h(x)=exp\left[\int^x a_0(s)ds + k_1\right] $$(2.14.1) $$\displaystyle y(x)=\frac {1}{h(x)}\left[\int^x h(s)b(s)ds + k_2 \right]$$(2.14.2) $$\displaystyle a_1(x)y'+a_0(x)y= b(x)$$(2.14.3)

Find
Solve the general L1-ODE-VC 1. $$\displaystyle a_1(x)=1$$ $$\displaystyle a_0(x)=x$$ $$\displaystyle b(x)=2x + 3 $$ 2. Find $$\displaystyle y(x)$$ in term of $$\displaystyle a_1(x), a_0(x), b(x)$$ 3. $$\displaystyle a_1(x)=x^2+1$$ $$\displaystyle a_0(x)=x$$ $$\displaystyle b(x)=2x$$

Solution
an integrating factor is a function that is chosen to facilitate the solving of a given equation involving differentials. It is commonly used to solve ordinary differential equations, but is also used within multivariable calculus, in this case often multiplying through by an integrating factor allows an inexact differential to be made into an exact differential. From lecture note, $$\underbrace{\left( hM \right)}_+\underbrace{\left( hN \right)}_{y}'=0$$ And applying second exactness condition, the equation is following that $$\underbrace{\left( h\frac{\partial M}{\partial y}+M\frac{\partial h}{\partial y} \right)}_{\frac{\partial \bar{M}}{\partial y}}=\underbrace{\left( h\frac{\partial N}{\partial x}+N\frac{\partial h}{\partial x} \right)}_{\frac{\partial \bar{N}}{\partial x}}$$ If we assume that $$\displaystyle h_y(x,y)=0$$, then h is function of x only and the equation is following that $$\left( h\frac{\partial M}{\partial y} \right)=\left( h\frac{\partial N}{\partial x}+N\frac{\partial h}{\partial x} \right)$$ And we obtain some equations through rearranging equation above $$\displaystyle h{{N}_{x}}+N{{h}_{x}}-h{{M}_{y}}=0$$ $$\displaystyle {{N}_{x}}=\frac{\partial N}{\partial x}$$ $$\displaystyle {{M}_{y}}=\frac{\partial M}{\partial y}$$ $$\displaystyle {{h}_{x}}=\frac{\partial h}{\partial x}$$ And we can obtain h(x) through dividing by h

$$\displaystyle \frac {h_x}{h}=-\frac{1}{N}(\underbrace{N_x}_{0}-\underbrace{M_y}_{a_0(x)})=a_0(x)$$ $$\displaystyle h(x)={{e}^{-\int{\frac{1}{N}\left( {{N}_{x}}-{{M}_{y}} \right)dx}}}=exp\left[\int^x a_0(s)ds + k_1\right] $$(2.14.1) And in order to obtain y(x), we can apply the integrating factor to obtain the following equations $$\displaystyle y'+a_0y=b \rightarrow h(y'+a_0y)=hb$$ and $$\displaystyle h_x/h=a_0 \rightarrow ha_0=h_x=h'$$ $$\displaystyle hy'+h'y=hb$$ Thus, $$\displaystyle y(x)=\frac {1}{h(x)}\left[\int^x h(s)b(s)ds + k_2 \right]$$(2.14.2)

part 1
1. In order to solve this problems, plug given information in eq (2.14.3) the equation is following that $$\displaystyle y'+ xy = 2x + 3 $$ Also, eq (2.14.1) is following that $$\displaystyle h(x)=exp\left[\int^x sds+k_1\right] $$ $$\displaystyle h(x)=exp\left[\frac {1}{2}s^2\right]$$ And eq (2.14.2) is following that $$\displaystyle y(x)=\frac {1}{h(x)}\left[\int^x h(s)b(s)ds + k_2 \right]$$ $$\displaystyle = \frac {1}{exp\left[\frac {1}{2}x^2\right]}\left[\int^x exp\left[\frac {1}{2}s^2\right]*(2s+3)ds+k_2\right]$$ $$\displaystyle =\frac {1}{exp\left[\frac {1}{2}x^2\right]}\left[\int^x e^{\frac {1}{2}s^2}*2sds+\int^x e^{\frac {1}{2}s^2}*3ds\right]$$ Let $$\displaystyle t= \frac {s^2}{2}$$ $$\displaystyle \frac {dt}{s}=ds$$ Finally, the equation is that $$\displaystyle y(x) =\frac {1}{exp\left[\frac {1}{2}x^2\right]}\left[\int^x 2e^tdt+\int^x e^{\frac {1}{2}s^2}*3ds\right]=\frac {1}{exp\left[\frac {1}{2}x^2\right]}\left[2e^{\frac {x^2}{2}}+\int^x e^{\frac {1}{2}s^2}*3ds\right]$$ In order to solve more, numerical method must be used.

part 2
From eq 2.14.3 $$\displaystyle a_1(x)y'+a_0(x)y= b(x)$$ $$\displaystyle y'+ \frac {a_0(x)}{a_1(x)}y=\frac {b(x)}{a_1(x)}$$ and plug in eq 2.14.1 $$\displaystyle h(x)=exp\left[\int^x \frac {a_0(s)}{a_1(s)}ds\right] $$ $$\displaystyle y(x)=\frac {1}{exp\left[\int^x \frac {a_0(s)}{a_1(s)}ds\right]}\left[\int^x exp \left[ \int^x \frac {a_0(s)}{a_1(s)}ds \right] \frac{b(s)}{a_1(s)}ds\right]$$