User:GatorEngineer13245

All assignments due at 5:00PM EST on the due date.

 Homework Set 1: (due Sept. 9) ---> (UPDATE - due Sept. 16th) Covers lectures from Aug. 24 through Sept. 4 

Problem 1: (Lecture 2-2, refers to 1-2)

Given a function $$f\left(y^1(t),t\right)$$, show that:

$$\frac{df}{dt}=\frac{\partial f}{\partial s}\dot{y}^1(t)+\frac{\partial f}{\partial t}$$

and

$$\frac{d^2f}{dt^2}=\frac{\partial f}{\partial s}\ddot{y}^1+\frac{\partial^2 f}{\partial^2 s}(\dot{y}^1)^2+2\frac{\partial^2 f}{\partial s \partial t}\dot{y}^1+\frac{\partial^2 f}{\partial^2 t}$$

Solution: Using the chain rule and the simplifying assumption that dy1=ds

$$\frac{df}{dt}=\frac{\partial f}{\partial s}\frac{\partial s}{\partial t}+\frac{\partial f}{\partial t}$$

ds/dt simplifies to $$\dot{y}^1$$ as shown

$$\frac{df}{dt}=\frac{\partial f}{\partial s}\dot{y}^1+\frac{\partial f}{\partial t}$$

For the second part we can write

$$\frac{d^2f}{dt^2}=\frac{d}{dt}\left ( \frac{df}{dt} \right )=\frac{d}{dt}\left (\frac{\partial f}{\partial s}\frac{\partial s}{\partial t}+\frac{\partial f}{\partial t} \right )$$

Using the product rule, this becomes:

$$\frac{d}{dt}\left (\frac{\partial f}{\partial s}\frac{\partial s}{\partial t}+\frac{\partial f}{\partial t} \right )=\frac{\partial s}{\partial t}\cdot \frac{d}{dt}\left (  \frac{\partial f}{\partial s} \right )+\frac{\partial f}{\partial s}\cdot \frac{d}{dt}\left ( \frac{\partial s}{\partial t} \right )+\frac{\partial^2 f}{\partial s \partial t}\cdot \frac{\partial s}{\partial t}+\frac{\partial^2 f}{\partial^2 t}$$

Completing the derivatives yields:

$$\frac{\partial s}{\partial t}\cdot \left ( \frac{\partial^2 f}{\partial^2 s}\cdot \frac{\partial s}{\partial t}+\frac{\partial^2 f}{\partial t \partial s} \right )+\frac{\partial f}{\partial s}\cdot \left ( \frac{\partial^2 s}{\partial^2 t} \right )+\frac{\partial^2 f}{\partial s \partial t}\cdot \frac{\partial s}{\partial t}+\frac{\partial^2 f}{\partial^2 t}$$

Again, substituting $$\dot{y}^1$$ for ds/dt and simplifying the expression yields

$$\frac{\partial^2 f}{\partial s^2}\left ( \dot{y}^1 \right )^2+2\cdot \frac{\partial^2 f}{\partial s \partial t}\dot{y}^1+\frac{\partial f}{\partial s}\ddot{y}^1+\frac{\partial^2 f}{\partial^2 t}$$

Problem 2: (Lecture 4-2)

Show that the solution of $$p^{\prime}+p=x$$ is $$p(x)=Ae^{-x}+x-1$$ using integrating factors. Problem 3: (Lecture 4-3)

Show that the first order ODE $$(2x^2+\sqrt{y})+x^5y^3y^{\prime}=0$$ is nonlinear

Solution A linear operator $$L\left ( \cdot \right )$$ is defined such that $$\forall\alpha,\beta \in\mathbb{R}$$, $$L(\alpha u+\beta v)=\alpha L(u)+\beta L(v)$$ Therefore, for $$F(x,y)=(2x^2+\sqrt{y})+x^5y^3y^{\prime}=0$$ to be linear Problem 4: (Lecture 5-2)

Given a function $$F(x,y,y^{\prime})=x^2y^5+6(y^{\prime})^2=0$$, show that F is a nonlinear 1st Order ODE. In other words, F does not satisfy the linearity condition $$\forall\alpha,\beta \in\mathbb{R}$$, $$F(\alpha u+\beta v)=\alpha F(u)+\beta F(v)$$ Problem 5: (Lecture 6-1)

Create an exact linear 1st Order ODE of the form $$\Phi_x(x,y)+\Phi_y(x,y)y^{\prime}=0$$ using the equation $$\Phi(x,y)=6x^4+2y^{3/2}$$ Create three more exact linear 1st Order ODEs by inventing new $$\Phi(x,y)$$ functions. Homework Set 2: (due Sept. 23) Covers lectures from Sept. 7 through Sept. 18 (not yet a complete listing) 

Problem 1: (Lecture 7-1)

Given $$h_xN-h_yM+h(N_x-M_y)=0$$, solve for h(x,y) given that $$h_xN=0$$ Problem 2: (Lecture 8-2)

Given the Non-homogeneous Linear 1st Order ODE with Varying Coefficients $$y^{\prime}+\frac{1}{x}y=x^2$$, verify that the integrating factor necessary to make the ODE exact is $$h(x)=x$$ Problem 3: (Lecture 9-2)

Show that the Linear 1st Order ODE-VC $$\frac{1}{2}x^2y^{\prime}+[x^4y+10]=0$$ is exact Problem 4: (Lecture 9-3)

Show that $$\frac{1}{3}x^3(y^4)y^{\prime}+(5x^3+2)(\frac{1}{5}y^5)=0$$ is an exact nonlinear 1st Order ODE

Problem 5: (Lecture 10-3)

Determine if the 2nd condition for exactness is satisfied for $$(xy)y{}''+x(y{}'^{2})+yy{}'=0$$