User:Egm6321.f10.team5.abc/2NDHW

Problem 1
Verify that $$M(x,y) + N(x,y) \frac {dy} {dx} = 0$$ statement, and give example for L1_ODE.

Solution 1
First, we let that $$M(x,y) + N(x,y) \frac {dy} {dx} = F(x,y,y')$$

Problem 2
Verify $$F(x,y,y') = M(x,y) + N(x,y)y' = (4x^7 + \sin y) + (x^2 y^3) y' = 0$$ is first-order nonlinear ODE.

Solution 2
We define a differential operator as $$D(\cdot) = (4x^7 + \sin (\cdot)) + x^2 (\cdot)^3 \frac {d(\cdot)} {dx} = 0$$ To suffice the condition of linearity, the following should be true $$D(\alpha u + \beta v) = \alpha D(u) + \beta D(v)$$ where $$\alpha$$ & $$\beta$$ are any arbitrary numbers and $$u$$ & $$v$$ are two possible arguments of $$D(\cdot)$$.

We derive $$D(\alpha u + \beta v)$$ first. $$D(\alpha u + \beta v) = [4x^7 + \sin (\alpha u + \beta v)] + x^2 (\alpha u + \beta v)^3 \frac {d(\alpha u + \beta v)} {dx}$$ $$= 4x^7 + \sin (\alpha u + \beta v) + x^2 (\alpha u + \beta v)^3 \left(\alpha \frac {du} {dx} + \beta \frac {dv} {dx}\right)$$ Now we derive $$\alpha D(u) + \beta D(v)$$. $$\alpha D(u) + \beta D(v) = \alpha [(4x^7 + \sin u) + x^2 u^3 \frac {du} {dx}] + \beta [(4x^7 + \sin v) + x^2 v^3 \frac {dv} {dx}]$$ Compare the two derivations and we evaluate that $$4x^7 + \sin (\alpha u + \beta v) + x^2 (\alpha u + \beta v)^3 \left(\alpha \frac {du} {dx} + \beta \frac {dv} {dx}\right) \ne \alpha [(4x^7 + \sin u) + x^2 u^3 \frac {du} {dx}] + \beta [(4x^7 + \sin v) + x^2 v^3 \frac {dv} {dx}]$$ Therefore we verified that the equation is first-order nonlinear ODE.

Problem 3
Plot $$y_H^1(x)$$ & $$y_H^2(x)$$ and show that $$y_H^1(x) \ne \alpha y_H^2(x)$$ for any $$x$$ and any $$\alpha$$ that belong to $$\mathbb{R}$$ where $$y_H^1(x) = x$$ and $$y_H^2(x) = \frac x 2 \log \left(\frac {1+x} {1-x}\right) - 1$$ $$\frac{x}{2} \log \left( \frac{1+1}{1-x} \right ) - 1$$

Solution 3
\includegraphics[scale=0.5]{homework} In the figure inserted above, the points of $$y_H^1(x)$$ are marked as "x" and the points of $$y_H^2(x)$$ are maked as "o". To find $$\hat x$$ such that $$y_H^1(\hat x) - \alpha y_H^2(\hat x) \ne 0$$ we rewrite the equation as following $$\hat x - \alpha \frac {\hat x} 2 \log \left(\frac {1+ \hat x} {1-\hat x}\right) - \alpha \ne 0$$ $$1 - \frac \alpha 2 \log \left(\frac {1+ \hat x} {1-\hat x}\right) \ne \frac 1 {\hat x}$$

Problem 4
Find $$F(x,y,y') = \frac {d\phi} {dx} (x,y)$$ where $$\phi(x,y) = x^2 y^{\frac 3 2} + \log (x^3 y^2) = k$$ and verify that $$F$$ is exact N1_ODE and invent 3 more.

Solution 4
First, we derive $$\frac {d\phi} {dx} (x,y)$$. $$F(x,y,y') = x y^{\frac 3 2} + \frac 1 {x^3 y^2} 3x^2 = x y^{\frac 3 2} + \frac 3 {x y^2}$$ The definition of exact N1_ODE for $$F$$ requires a function $$\phi(x,y)$$ such that $$F = \frac {d\phi} {dx}$$ to exist. Since we have the function $$\phi(x,y)$$, it is verified that F is exact N1_ODE. The following three functions are also exact N1_ODE.

1. $$F(x,y,y') = 2x$$. Then, we have a function $$\phi(x,y) = x^2$$ such that $$F = \frac {d\phi} {dx}$$

2. $$F(x,y,y') = \frac 1 x$$. Then, we have a function $$\phi(x,y) = \log(x)$$ such that $$F = \frac {d\phi} {dx}$$

3. $$F(x,y,y') = 1$$. Then, we have a function $$\phi(x,y) = x$$ such that $$F = \frac {d\phi} {dx}$$

Problem 5
$$M(x,y)+N(x,y)f(y')=0$$. Find $$f(y')$$ such that there is no analytical solution to $$f(y')=-\frac M N$$ which means such N1_ODE cannot be exact.

Solution 5
The equation has the form to satisfy the first condition of exactness. So, we need to find $$f(y')$$ that cannot satisfy the second condition of exactness which is $$M_y(x,y) = N_x(x,y)$$.

Problem 6
Verify $$y(x) = \sin^{-1}(k-15x^5)$$ satisfy $$M(x,y)+N(x,y)y'=75x^4 + \cos y \ y'=0$$

Solution 6
First, transform the $$y = \sin^{-1}(k-15x^5)$$ and express as following. $$\sin y = k - 15x^5$$ Then, take a derivative of the new expression with respect to $$x$$. $$\cos y \ y' = - 75 x^4$$ Now we verify that $$y$$ satisfies $$M(x,y)+N(x,y)y'$$ $$M(x,y)+N(x,y)y'=75x^4 - 75x^4 = 0$$

Problem 7
Solve $$a_1(x) y' + a_0(x) y = b(x)$$ where $$a_0(x) = x$$, $$a_1(x) = 1$$, and $$b(x) = 2x+3$$. Do the same for $$a_0(x) = x$$, $$a_1(x) = x^2+1$$, and $$b(x) = 2x$$. Lastly, assume that $$a_1(x) \ne 0$$ for any $$x$$ becomes $$y' + \frac {a_0(x)} {a_1(x)} y = \frac {b(x)} {a_1(x)}$$ and find the expression for $$y(x)$$ in terms of $$a_0$$, $$a_1$$, and $$b$$.

Solution 7
We rewrite equation with all the constants for the case of $$a_0(x) = x$$, $$a_1(x) = 1$$, and $$b(x) = 2x+3$$. $$y' + x y = 2x + 3$$ Since the form of $$hy'+h'y = (hy)'$$ is needed, multiply the equation by $$h$$. $$hy' + hxy = h (2x + 3)$$ From the last equation, we have $$hx = h' = h_x$$. $$\frac {h_x} h = x$$ We could also derive $$\frac {h_x} h$$ by $$\frac {h_x} h = - \frac 1 N (N_x - M_y)$$ Because the equation is in the form of $$M(x,y) + N(x,y)y' = 0$$, $$[M(x,y) = xy-2x-3$$ and $$N(x,y) = 1$$. $$\frac {h_x} h = - \frac 1 1 (0 - x) = x$$ $$\frac 1 h dh = x dx$$ $$h = \exp {\frac {x^2} 2}$$ As we can make the equation of $$(hy)' = hb$$ form, we solve the equation by $$y(x) = \frac 1 {h(x)} \int^x h(s) b(s) ds = \frac 1 {\exp {\frac {x^2} 2}} \int^x \exp {\frac {s^2} 2} (2s+3) ds = 3 \sqrt 2 \ \mbox{F}(\frac x {\sqrt 2}) + 2$$ where F(x) is Dawson function. Now consider the other case where $$a_0(x) = x$$, $$a_1(x) = x^2+1$$, and $$b(x) = 2x$$. $$(x^2+1) y' + x y = 2x$$ $$\frac {h_x} h = - \frac 1 N (N_x - M_y) = - \frac 1 {(x^2+1)} (2x - x) = - \frac x {x^2+1}$$ $$\frac 1 h dh = - \frac x {x^2+1} dx$$ $$\ln h = - \frac 1 2 \ln(x^2+1)$$ $$h = (x^2+1)^{- \frac 1 2}$$ We solve for $$y(x)$$ at this time by integration. $$y(x) = \frac 1 {h(x)} \int^x h(s) b(s) = \frac 1 {(x^2+1)^{- \frac 1 2}} \int^x (s^2+1)^{- \frac 1 2} 2s ds = \sqrt{x^2+1} \ 2\sqrt{x^2+1} = 2 (x^2+1)$$ Lastly, we find the expression for y(x) in the event of $$y' + P(x) y = Q(x)$$ where $$P(x) = \frac {a_0(x)} {a_1(x)}$$ and $$Q(x) = \frac {b(x)} {a_1(x)}$$. $$\frac {h_x} h = - \frac 1 N (N_x - M_y) = P(x)$$ $$h(x) = \exp[\int^x P(s) ds] = \exp[\int^x \frac {a_0(s)} {a_1(s)} ds]$$ $$y(x) = \frac 1 {h(x)} \int^x h(s) Q(s) ds = \frac 1 {\exp[\int^x \frac {a_0(s)} {a_1(s)} ds]} \int^x \exp[\int^s \frac {a_0(r)} {a_1(r)} dr] \frac {b(s)} {a_1(s)} ds$$

Problem 8
First, show that $$k_1$$ is not necessary. Then show that $$y(x) = \frac 1 {h(x)} \int^x h(s) b(s) ds$$ agrees with $$y(x) = Ay_H(x) + y_P(x)$$. Finally, find $$y_H(x)$$ independent which means solving $$y'+a_0 y = 0$$

Solution 8
First, we have a constant term from $$\int^x_{} a_0(s) ds$$ so that $$h(x) = \exp[\int^x a_0(s) ds] = \exp[f(x) + k_1] = \exp f(x) \cdot \exp k_1$$ where $$f(x)$$ is the antiderivative and $$k_1$$ is the constant. If we substitute this expression inthe the equation of $$y(x)$$, we see that $$y(x) = \frac 1 {h(x)} \int^x h(s) b(s) ds = \frac 1 {\mbox{e}^{f(x)} \mbox{e}^{k_1}} \int^x \mbox{e}^{f(s)} \mbox{e}^{k_1} b(s) ds$$ Since $$\mbox{e}^{k_1}$$ is constant, it is possible cancel the two constants in the denominator and in the numerator. Therefore $$k_1$$ is not necessary. \\ Similarly to the first step, we also have a constant term from $$y(x) = \frac 1 {h(x)} \int^x h(s) b(s) ds$$ such that $$y(x) = \frac 1 {h(x)} \int^x h(s) b(s) ds = \frac 1 {h(x)} [f(x) + k_2] = \frac {f(x)} {h(x)} + \frac {k_2} {h(x)}$$ where $$f(x)$$ is the antiderivative and $$k_2$$ is the constant. Compare this equation to $$y(x) = A y_H(x) + y_P(x)$$ We see that $$\frac {f(x)} {h(x)} = y_P(x)$$ and $$\frac {k_2} {h(x)} = A y_H(x)$$. Therefore, the two equations agree with each other. Finally, we solve $$y' + a_0 y = 0$$ to obtain a homogeneous solution independently. $$\frac {y'} y = - a_0(x)$$ $$\frac {dy} y = - a_0(x) dx$$ $$\ln y = - \int^x_{} a_0(s) ds$$ $$y = \exp(- \int^x_{} a_0(s) ds)$$

Problem 9
Use $$\bar b(x) c(y) y' + a(x) \bar c(y) = 0$$ to find a N1_ODE that is either exact or can be made exact where $$a(x) = \sin x^3$$, $$b(x) = \cos x$$, and $$c(y) = \exp(2y)$$. Then find $$\phi(x,y) = k$$.

Solution 9
First, calculate $$\bar b(x)$$ and $$\bar c(x)$$. $$\bar b(x) = \int^x b(s) ds = \int^x \cos s \ ds = \sin x$$ $$\bar c(x) = \int^y c(y) ds = \int^y \exp(2s) ds = \frac 1 2 \exp(2y)$$ Thus a N1_ODE that is either exact or can be made exact is $$\sin x \ \frac 1 2 \exp(2y) y' + \sin x^3 \ \frac 1 2 \exp(2y) = 0$$ To find $$\phi(x,y) = k$$, we use the following relation. $$F(x,y,y') = \frac {d\phi(x,y)} {dx} = M(x,y)$$