1st order linear equations

In summary, the conversation is about solving a differential equation in the form of dy/dx + P(x)y = Q(x). The person is struggling with converting the given equation (2e^y -x) dy/dx = 1 into this form and is seeking help. Another equation is also given, (x + y^2)dy = ydx, which the person is also having trouble converting. The person is then given a hint to use the integrating factor method and is guided through the steps to solve the first equation. However, there is some confusion about the notation and the person is unsure how to solve the integral in the second equation. They also mention that their answer does not match the one in the back of the
  • #1
ch2kb0x
31
0

Homework Statement



(2e^y -x) dy/dx = 1

Homework Equations



dy/dx + P(x)y = Q(x)

The Attempt at a Solution



I know how to solve these equations, but I can't get this into the dy/dx + P(x)y = Q(x) form.

In addition to this problem, is also this: (x + y^2)dy = ydx (can't get into the form). Help would be appreciated.

Also, for both of these questions, they are asking to consider x as function of x = x(y).
 
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  • #2
ch2kb0x said:

Homework Statement



(2e^y -x) dy/dx = 1


Also, for both of these questions, they are asking to consider x as function of x = x(y).

Use the hint, using dy/dx = 1 / (dx/dy) you can rewrite it as:

dx/dy + x = 2ey

It looks strange because you are used to thinking of y as a function of x. If the x and y were switched it would look familiar:

dy/dx + y = 2ex

You could even switch them and work it on a sheet of scratch paper, then copy it back with the x's and y's switched back if that helps you visualize it.
 
  • #3
So, if I were to solve, would it be like this:

dx/dy + x = 2e^y

I(x) = e^integral dy = e^y

multiply integrating factor on both sides will give: e^y (x) = e^(2y) + C

=> x = e^y + Ce^-y.

That was the answer I got for x, but it does not match answer in the back.

help.
 
  • #4
ch2kb0x said:
So, if I were to solve, would it be like this:

dx/dy + x = 2e^y

I(x) = e^integral dy = e^y

multiply integrating factor on both sides will give: e^y (x) = e^(2y) + C

=> x = e^y + Ce^-y.

That was the answer I got for x, but it does not match answer in the back.

help.

It's hard to get used to switching the independent and dependent variables y and x. Your integrating factor isn't I(x) now, it's I(y). Your solution looks OK other than that notational thing. And, regardless what the solution manual may say, you can check that your version of the solution solves the above equation.
 
  • #5
Hmmm ok, still kind of confused. this is what I did after your correction:

dx/dy + x = 2e^y

I(y) = e^integral (2e^y) dy = e^(2e^y)

multiply integrating factor on both sides will give: e^(2e^y) x = Integral e^(2e^y) (2e^y) dy...i am unsure on how to solve that integral on the right side.
 
  • #6
ch2kb0x said:
Hmmm ok, still kind of confused. this is what I did after your correction:

dx/dy + x = 2e^y

I(y) = e^integral (2e^y) dy = e^(2e^y)

multiply integrating factor on both sides will give: e^(2e^y) x = Integral e^(2e^y) (2e^y) dy...i am unsure on how to solve that integral on the right side.

No no. I was just pointing out that your integrating factor is a function of y, not x. As I said before, the solution you had works. You don't need to change it.
 
  • #7
ch2kb0x said:
So, if I were to solve, would it be like this:

dx/dy + x = 2e^y

I(x) = e^integral dy = e^y

multiply integrating factor on both sides will give: e^y (x) = e^(2y) + C

=> x = e^y + Ce^-y.

That was the answer I got for x, but it does not match answer in the back.

help.
What is the answer in the book? Knowing that might help us!
 

1. What is a 1st order linear equation?

A 1st order linear equation is a mathematical expression that involves one variable raised to the first power, with a coefficient and a constant. It is in the form of y = mx + b, where m is the slope and b is the y-intercept.

2. How do you solve a 1st order linear equation?

To solve a 1st order linear equation, you can use the slope-intercept form (y = mx + b) or the point-slope form (y - y1 = m(x - x1)). Plug in the given values for m, x, and y and solve for the remaining variable. You can also use algebraic methods such as elimination or substitution to solve the equation.

3. What is the difference between a linear and a non-linear equation?

A linear equation has a constant rate of change, meaning that the graph of the equation will form a straight line. On the other hand, a non-linear equation does not have a constant rate of change and the graph will not form a straight line.

4. Can a 1st order linear equation have more than one solution?

Yes, a 1st order linear equation can have infinitely many solutions. This occurs when the equation has the same slope and y-intercept as another line, resulting in all points on the two lines being solutions to the equation.

5. How are 1st order linear equations used in real life?

1st order linear equations are commonly used in various fields such as physics, economics, and engineering to model and solve real-life problems. They can be used to calculate rates of change, predict future values, and analyze data trends.

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