Solve $\displaystyle 3y'-2y=e^{-\pi/2}\quad y(0)=a$: Find $a_0$ Exactly

• MHB
• karush
In summary, the conversation discusses solving an initial value problem and finding the critical value $\displaystyle a_0$ exactly. The process involves dividing by 3, using an integrating factor, and integrating both sides of the equation. Alternatively, the problem can also be solved by solving the characteristic equation and finding the general solution. The value of the undetermined constant can be determined using the initial condition.
karush
Gold Member
MHB
\tiny{b.2.1.23}
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Solve the initial value problem and find the critical value $\displaystyle a_0$ exactly.
$\displaystyle 3y'-2y=e^{-\pi/2}\quad y(0)=a$
divide by 3
$\displaystyle y'-\frac{2}{3}y=\frac{e^{-\pi/2}}{3}$
with integrating factor $\displaystyle e^{-2t/3}$ multiply through
$\displaystyle e^{-2t/3}y'-\frac{2}{3} e^{-2t/3}y=\frac{e^{-\pi/2}}{3}e^{-2t/3}$
ok just seeing if this is going in the right direction

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It's easy to check. If $e^{-2t/3}$ is an integrating factor then $\frac{d}{dt}(e^{-2t/3}y)= e^{-2t/3}y'- \frac{2}{3}e^{-2t/3}y= e^{-2t/3}\frac{e^{-\pi/2}}{3}$ so, yes, that is the correct integrating factor.To complete, integrate both sides of $\frac{d}{dt}\left(e^{-2t/3}y\right)= e^{-(4t- 3\pi)/6}$ which is, of course, the same as $d\left(e^{-2t/3}y\right) =\left(e^{-(4t- 3\pi)/6}\right)dt$ including the constant of integration, using the condition that y(0)= 1 to determine the value of that constant.

Since this is a "linear equation with constant coefficients" it can also be done, perhaps more easily, by solving the "characteristic equation", $3r- 2= 0$ so $r= \frac{2}{3}$ meaning that the general solution to the "associated homogeneous equation", 3y'- 2y= 0, is $y_h(t)= Ce^{2t/3}$ where C is an undetermined constant. We then look for a single function to satisfy the entire equation $3y'- 2y= e^{-\pi/2}$. Since the right side is a contant, we try y= A so that y'= 0 and the equation becomes $-2A- e^{-\pi/2}$ so that $A= -\frac{1}{2}e^{-pi/2}$ and the general solution to the entire equation is $y(t)= Ce^{2t/3}- \frac{1}{2}e^{-\pi/2}$. Use $y(0)= C- \frac{1}{2}e^{-\pi/2}= 1$ to determine the value of C.

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1. What is the meaning of the equation $\displaystyle 3y'-2y=e^{-\pi/2}$?

The equation represents a first-order linear differential equation, where the derivative of the function y is multiplied by a constant and then subtracted from the function itself. The right side of the equation contains the constant e raised to the power of -π/2.

2. What does the initial condition $y(0)=a$ represent?

The initial condition represents the value of the function y at the point x=0. In other words, it gives a specific starting point for the function to be solved.

3. How do you solve this differential equation?

To solve this differential equation, we can use the method of integrating factors. First, we rearrange the equation to the form y' + p(x)y = q(x), where p(x) = -2/3 and q(x) = e^(-π/2)/3. Then, we find the integrating factor u(x) = e^(∫p(x)dx) = e^(-2/3x). Multiplying both sides of the equation by u(x) and integrating, we get y(x) = e^(2/3x)∫e^(-π/2)/3e^(-2/3x)dx + Ce^(2/3x). Finally, applying the initial condition, we can solve for the constant C and get the exact solution.

4. What is the significance of finding $a_0$?

Finding $a_0$ allows us to determine the exact solution to the differential equation. It represents the value of the function y at the point x=0, which is the initial condition given in the problem. By finding this value, we can solve for the constant C and have a complete solution to the equation.

5. Can this differential equation be solved using other methods?

Yes, there are other methods that can be used to solve this differential equation, such as the method of variation of parameters or using a power series solution. However, the method of integrating factors is the most commonly used and efficient method for solving first-order linear differential equations.

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