Homework 2b
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Homework 2b:
Read the following and turn in on paper answers to the 3 problems below that are identified as homework problems (some problems have multiple parts):1.2 = 2.2: Separation of variables.
A very useful method to solve mathematics problems is to convert a problem into one you already know how to solve. To solve the differential equations in section 1.2 and 2.2, you can covert these problems into single variable calculus integration problems. Example: $~~x \frac{dy}{dx} = y$ To make this look more like an integration problem from calculus, we will multiply both sides by $dx$ since integration problems do not have $dx$ (or $dy$) in the denominator: xdy=ydx But $xdy$ (and $ydx$) are not single variable calculus problems since they both involve 2 variables. Thus we need to separate the variables: dyy=dxx Note to separate the variables, I moved the $x$ to the side with the $dx$ by dividing both sides by $x$ and the $y$ to the side with the $dy$ by dividing both sides by $y$. I did not want to move $dx$ or $dy$ since they are already on opposite sides, and I don't want to divide by either of them since we don't want $dx$ or $dy$ in the denominator. To get rid of the derivative, we can integrate both sides: ∫dyy=∫dxx Both the LHS and RHS are now single variable calculus 1 problem. Thus we have changed our original differential equation into a problem we know how to do. After integrating: ln|y|=ln|x|+C We now have an algebra problem where we can solve for $y$: eln|y|=eln|x|+C=eln|x|eC=|x|eC=xC Thus $y = Cx$ since $e^C$ is a constant. FYI: In the above I lost solution the solution $y = 0$ when I divided by $y$, but I regained it when I replaced $e^C$ with $C$. In the real world you should be careful about losing solutions and replacing a positive number like $e^C$ with an arbitrary constant $C$. We won't be that careful in this class. HW 2b problem 1: Use separation of variables to solve $y' = x^2y + y$ Hint 1: To separate the variables, you need to see all the variables, so rewrite $y'$ as $\frac{dy}{dx}$ Hint 2: To separate the variables, $x$ from $y$, sometimes you will need to factor. For example, in the above on the RHS, you can factor out the $y$ before moving it to the other side (using the appropriate algebraic operation).
2.1: Solving linear differential equations using an integrating factor.
A first order differential equation is linear if it can be written in the form: $a(t)y' + b(t)y = g(t)$. It is called "linear" since the LHS of the equation is a linear combination of $y$ and $y'$ (where the coefficients are functions of $t$) and there are no $y$'s (including $y'$'s) on the RHS. Simple motivating example: $~~e^{3t}y' + 3e^{3t}y = 1$ Note the LHS comes from the product rule: $~~e^{3t}y' + 3e^{3t}y = [e^{3t}y]'$ Thus we have: $~~[e^{3t}y]' = 1$ We can now integrate both sides using the Fundamental Theorem of Calculus: $\int F'(t)dt = F(t) +C$. In other words the integral of the derivative of a function returns the function, offset by a constant $C$. Integrating both sides: $~~\int[e^{3t}y]'dt = \int 1dt$ Hence $~~e^{3t}y = t + C$ and thus $y = te^{-3t} + Ce^{-3t}$ Please avoid common errors: Remember the $+C$ and if you divide one side by $e^{3t}$, you must divide the entire other side (both the $t$ and the $C$ by $e^{3t}$). Real example: $~~ty' + 2y = \frac{e^t}{t}$ We want to create the product rule: $u(t)y' + u'(t)y = [u(t) y]'$ since $[u(t) y]'$ is easy to integrate using the FTC. To do so, we will use an integrating factor. For the differential equation $y' + p(t)y = g(t)$, the integrating factor is $u(t) = e^{\int p(t)dt}$ Optional thought question: After reading the below, do you know why we chose $u(t) = e^{\int p(t)dt}~$? Step 0: Divide both sides by $t$ since to use the above formula for the integrating factor, the coefficient in front of $y'$ must be a one. 1y′+2ty=ett2 Step 1: Find the integrating factor: The coefficient of $y$ is $p(t) = \frac{2}{t}$ and thus the integrating factor is $u(t) = e^{\int p(t)dt}$ is u(t)=e∫p(t)dt=e∫2tdt=e2ln(t)=eln(t2)=t2 Note in the above, I was sloppy: I did not use absolute value signs or the +C since I just need one integrating factor that works. We also used the log rule: $~~~ln(a^r) = r ln(a)$.We had to use this log rule to move the 2 as we needed the exponential function applied directly to the logarithmic function for these functions to cancel each other out. Step 2: Multiply both sides of the differential equation by the integrating factor $u(t) = t^2$. y′+2ty=ett2 t2y′+2ty=et You should now see the product rule on the LHS: [u(t)y]′=[t2y]′=t2y′+2ty Thus $~~~~~~~~t^2y' + {2}{t}y = e^t$ simplifies to $~~~[t^2y]' = e^t$ We can now integrate both sides, using the fundamental theorem of calculus to easily integrate the RHS. In this case I have fixed the problem to make it easy to also integrate the LHS, but often integrating the LHS is the most time-consuming part of the problem -- make sure you know how to do integration by substitution, by parts, and by partial fractions. ∫[t2y]′=∫et t2y=et+C We can now easily solve for $y$ by dividing both sides by the integrating factor $u(t) = t^2$. y=ett2+Ct2 HW 2b problem 2: Solve the following differential equation. Clearly indicate the step where you use the product rule ty′+3y=4t3(t+2) Hint 1: Sometimes (but not always) you will need to use a log rule to simplify the integrating factor. For this problem you will need to use a log rule. Hint 2: We will use partial fractions in chapter 6, so make sure you remember how to do integration by partial fractions. This problem involves integration by partial fractions. Hint for one of the 2.1 WP homework problems: To find the limit, you do not need to integrate the RHS. Instead, you can use l'hopital's rule. Hint for one of the 2.3 WP homework problems: To find the amplitude of $Acos(wt) + Bsin(wt)$, use that $Acos(wt) + Bsin(wt) = Rcos(wt - \delta)$ where $R = \sqrt{A^2 + B^2}$. FYI: This formula comes from converting $(Acos(wt), Bsin(wt))$ into polar coordinates $(R, \delta)$ where $R$ is the radius of the circle centered at (0, 0) and containing the point $(Acos(wt), Bsin(wt))$. We will use this trig formula in sections 3.7 and 3.8 when we discuss mechanical and electrical vibrations.