Linear and Bernoulli First Order

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The discussion focuses on solving three differential equations, identifying the first as a Linear Differential Equation (LDE) and the second and third as Bernoulli's Differential Equations (BDE). The LDE follows the format dy/dx + yP(x) = Q(x), while BDE includes a term Q(x)y^n. The user initially struggled with the correct identification of P(x) and Q(x) in the first problem, which they later determined could also be solved using Bernoulli's method. After modifications, they successfully solved all three equations, confirming that the first and third problems could be approached as LDEs, yielding consistent results with provided answers. The user concludes that the equations were misclassified but ultimately resolved correctly.
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1. The problem statement.

1. http://img194.imageshack.us/img194/3815/linear.png

2. http://img25.imageshack.us/img25/1543/bernoulli1.jpg

3. http://img23.imageshack.us/img23/7991/bernoulli2.jpg

2.Relevant equations

Problem 1 is Linear Differential Equation and Problems 2 and 3 are Bernoulli's D.E.

LDE is given by: dy/dx + yP(x) = Q(x)
BDE is given by: dy/dx + yP(x) = Q(x)y^n

The Attempt at a Solution



On those three equations, I'm trying to make the equations as exactly to the format on the LDE and BDE. However, I cannot get P(x) and Q(x) on no.1. The y is always y^3. In no.2, there will be Q(x)y^n and there is no y. In no.3, I solved that. However, that problem is BDE and I solved it using LDE with dx/dy. Do you think there is something wrong with the equation?

Thank you.
 
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I think I solved the three equations. There's something wrong with those equations. The first problem can be solved by using Bernoulli but it is under the linear problem set. I got the same answer exactly as the provided answer. The second problem is Bernoulli. I modified the equation and when I solved it, I got the same answer exactly as the provided answer. The equation goes 2x^3y'=y(y^2+3x^2). The third problem is the same as the first. Instead of Bernoulli, it is solvable using Linear. Anyway, thank you.
 
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