Solving 3y' - 2y = 1 - x and 2y' + 3y = 4 with initial condition

[primus]

Hi, I need help to solve two differential equation:

1. Find all solutions to the differential equation 3y' - 2y = 1 - x

2. Find the solution to 2y' + 3y = 4 when y(2) = 0

I would be happy if anyone could explain the general rule to solve these two equations, because the book I use only show poor examples.

 

[saltydog]

Hi, I need help to solve two differential equation:

1. Find all solutions to the differential equation 3y' - 2y = 1 - x

2. Find the solution to 2y' + 3y = 4 when y(2) = 0

I would be happy if anyone could explain the general rule to solve these two equations, because the book I use only show poor examples.

Hello Primus,

These are first-order ODEs: The technique is to find an integrating factor, multiply both sides by it, then integrate. The first case, integrate over a general interval ($x_o$,x), in the second case, integrate over a definite interval (2,x). However, you may not be familiar with these operations. I'll give you some time to look into them on your own. If you still have questions, ask, and me or someone else will work through them with you.

 

[HallsofIvy]

Integrating factor:
In problem 1, find a function u(x) so that 3(u(x)y(x))'= 3u(x)y'- 2u(x)y
Multiply the entire equation by u(x) and the left side is easy to integrate.

In problem 2, find a function u(x) so that 2(u(x)y(x))'= 2u(x)y'+ 3u(x)y.

 

[dextercioby]

Linear nonhomogenous constant coeff.I-st order ODE-s.

1.Solve the homogenous equation bby separating variables.
2.Find the particular solution to the nonhomogenous equation (through Lagrange's method).
3.Write the general solution & impose the initial condition (for the second which is a Cauchy problem).


Daniel.

 

[saltydog]

Well, here's the first step:

Whenever you have a first order ODE in that form, place it in standard form:

y'+A(x)y=B(x)

Thus, you have:

$$y^{'}-\frac{2}{3}y=\frac{1}{3}(1-x)$$

Once you have it in standard form, the integration factor is e raised to the integral of A(x):

$$\mu(x)=e^{\int -\frac{2}{3}dx}=e^{-\frac{2}{3}x$$

Now, just multiply both sides of the standard form equation by the intgration factor:

$$e^{-\frac{2}{3}x}(y'-\frac{2}{3}y)=\frac{1}{3}e^{-\frac{2}{3}x}(1-x)$$

The left side is the differential of $e^{-2/3 x}y$ so when you integrate that, you're left with just $e^{-2/3 x}y$ right?

Just integrate indefinitely both sides now and remember to add a constant of integation. It's easier than figuring it by integrating from $x_0$ to x.