# Thread: First-Order ODE

1. ## First-Order ODE

I have an assignment due on monday and I am having trouble. It deals with first order differential equations. I have a grasp of the methods, but am having trouble with algebraic manipulation to get the problems into the right form, at least that is where I think my problem lies. If anyone could help I would appreciate it. Even if you could just give me a hint (like think about completing the square) that would be nice.
I solved the first two problems already.

The assignment is attached in a pdf.

2. ## re: First-Order ODE

for problem three i've got it worked out this far.

3. ## re: First-Order ODE

Originally Posted by junwagh
for problem three i've got it worked out this far.
For question 3 I would start by dividing numerator and denominator by y^2, then making the substitution v = y/x.

4. ## re: First-Order ODE

Hey that's the first thing I tried but the variables on the right side are not in y/x form so I can't substitute.

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5. ## re: First-Order ODE

Originally Posted by junwagh
Hey that's the first thing I tried but the variables on the right side are not in y/x form so I can't substitute.

[IMG][/IMG]
You should know that x/y = (y/x)^(-1), and also if v = y/x then y = vx...

6. ## re: First-Order ODE

Thank you for that last tip. I substituted the v in, but I have to say I am stuck again. I have no idea how to solve the resulting equation. I don't think it can be an exact question and I can't get it into linear form to find an integrating factor. The variables don't seem separable either. I dunno what to do. I multiplied the right side by v^2/v^2 but am still stuck trying to figure out waht to do.

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7. ## re: First-Order ODE

Originally Posted by junwagh
Thank you for that last tip. I substituted the v in, but I have to say I am stuck again. I have no idea how to solve the resulting equation. I don't think it can be an exact question and I can't get it into linear form to find an integrating factor. The variables don't seem separable either. I dunno what to do. I multiplied the right side by v^2/v^2 but am still stuck trying to figure out waht to do.

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\displaystyle \displaystyle \begin{align*} x\,\frac{dv}{dx} + v &= \frac{v^{-2} + 1}{v\,x - 2v^{-1}} \\ x\,\frac{dv}{dx} + v &= \frac{1 + v^2}{v^3x - 2v} \\ x\,\frac{dv}{dx} &= \frac{1 + v^2}{v^3x - 2v} - v \\ x\,\frac{dv}{dx} &= \frac{1 + v^2}{v^3x - 2v} - \frac{v(v^3x - 2v)}{v^3x - 2v} \\ x\,\frac{dv}{dx} &= \frac{1 + v^2}{v^3x - 2v} - \frac{v^4x - 2v^2}{v^3x - 2v} \\ x\,\frac{dv}{dx} &= \frac{1 + 3v^2 - v^4x}{v^3x - 2v} \end{align*}

I need to think this one over...

4.
\displaystyle \displaystyle \begin{align*} \frac{dy}{dx} &= \frac{1}{e^{4y}+2x} \\ \frac{dx}{dy} &= e^{4y}+2x \\ \frac{dx}{dy} - 2x &= e^{4y} \end{align*}

This is now first-order linear.

5.
\displaystyle \displaystyle \begin{align*} x\,\frac{dy}{dx} + x\,y &= 1 - y \\ \frac{dy}{dx} + y &= \frac{1}{x} - \left(\frac{1}{x}\right)y \\ \frac{dy}{dx} + y + \left(\frac{1}{x}\right)y &= \frac{1}{x} \\ \frac{dy}{dx} + \left(1 + \frac{1}{x}\right)y &= \frac{1}{x} \end{align*}

This is now first-order linear.

8. ## re: First-Order ODE

Thank you so much for your help so far. This is much appreciated!!!

9. ## re: First-Order ODE

I figured out number three. The original problem can be converted into an exact equation I think.