Solve the differential equation:
dy/dx = y^2 + y
dy/(y^2+y) = dx
but 1/(y^2+y) = 1/(y-1) - 1/y, so
(1/(y-1) - 1/y)dy = dx
ln(y-1) - ln y = x+k
ln ((y-1)/y) = x+k
(y-1)/y = e^(x+k) = e^k * e^x = ce^x
y-1 = cye^x
y(1-ce^x) = 1
y = 1/(1-ce^x)
Hmmm. Wolframalpha says
y = ce^x/(e^cx-1)
but they are the same, since
u/(u-1) = 1 + 1/(u-1) so there's just a different constant c. Still, better double-check my math.
To solve the differential equation dy/dx = y^2 + y, we can use separation of variables.
1. Start by separating the variables, which means moving all the terms with y on one side and all the terms with x on the other side.
dy/(y^2 + y) = dx
2. Now, we can integrate both sides with respect to their respective variables.
∫dy/(y^2 + y) = ∫dx
3. The left side integral can be solved by using the method of partial fractions.
We can factor the denominator: y^2 + y = y(y + 1).
∫ [A/y + B/(y + 1)] dy = x + C
4. Find the values of A and B by equating the numerators.
A(y + 1) + By = 1.
Putting y = 0, we get A = 1.
Putting y = -1, we get -B = 1, which means B = -1.
∫dy/y - ∫dy/(y + 1) = x + C
5. Simplify and solve the integrals.
ln|y| - ln|y + 1| = x + C
6. Combine the logarithms using the properties of logarithms.
ln|y/(y + 1)| = x + C
Taking the exponential of both sides:
|y/(y + 1)| = e^(x + C)
Note: The absolute values cancel out since the exponential function is always positive.
7. Remove the absolute value and simplify.
y/(y + 1) = e^(x + C)
8. Solve for y.
Multiply both sides by (y + 1):
y = e^(x + C)(y + 1)
9. Distribute e^(x + C) to get:
y = e^x * e^C * (y + 1)
Let D = e^C, then we have:
y = D * e^x * (y + 1)
10. Finally, rearrange the equation to isolate y:
y - D * e^x * y = D * e^x
Factor out y:
y(1 - D * e^x) = D * e^x
Divide both sides by (1 - D * e^x):
y = D * e^x / (1 - D * e^x)
That's the solution to the differential equation dy/dx = y^2 + y. The constant D represents the initial condition, which can be determined if more information or specific conditions are given.