Simple differential equation question

[thursday]

is there anyone who can help me.

how do I solve the following differential equation:

dP = P^2V
dV

im not sure what the intergral of 1/P^2 is


is the answer P = Ce^(v^2) - Ce to the power of v squared ?
thanx

 

[HallsofIvy]

1/P²= P^-2. Do you know the integral of that?

"is the answer P = Ce^(v^2) - Ce to the power of v squared ?"

No, it's not anything like that. For one thing, it does not involve any exponentials.

 

[M98Ranger]

Sure, I can help you with this differential equation question. To solve it, we can use the method of separation of variables. First, let's rewrite the equation as:

dP/dV = P^2V

Next, we can separate the variables by moving all the terms with P to one side and all the terms with V to the other side:

dP/P^2 = VdV

Now, we can integrate both sides:

∫dP/P^2 = ∫VdV

To find the integral of 1/P^2, we can use the power rule for integration, which states that ∫x^n dx = x^(n+1)/(n+1) + C. In this case, n = -2, so the integral of 1/P^2 becomes P^-1 + C.

Substituting this into our equation, we get:

P^-1 + C = V^2/2 + C

Next, we can solve for P by isolating it on one side:

P^-1 = V^2/2 + C - C

P^-1 = V^2/2

Finally, we can take the reciprocal of both sides to get the solution for P:

P = 2/V^2

So the solution to the differential equation is P = 2/V^2 + C. I'm not sure where you got the answer P = Ce^(v^2) - Ce^(v^2) from, but it does not appear to be the correct solution. I hope this helps!