How would I use separation of variables to solve the initial value problem?
dy/dx = (y +5)(x+2)
I multiply both sides by dx so: dy= (y+5)(x+2)dx
Then I get variables on the same sides:: dy/(y+5)= (x+2)dx
Then I have to take integrals of both sides but this is where I get stuck. Can someone show me the rest? Thanks.
what is wrong with
ln(y+5)=1/2 (x+2)^2 +C
check
1/(y+5)dy=(x+2)dx
dy/dx=(x+2)(y+5)
To continue solving the initial value problem using separation of variables, you need to integrate both sides of the equation. Here's how you can do it:
∫ dy / (y + 5) = ∫ (x + 2) dx
To integrate the left side, you can use the substitution method. Let u = y + 5, then du = dy. The equation becomes:
∫ du / u = ∫ (x + 2) dx
The integral of 1/u with respect to u is ln|u| + C1, where C1 is the constant of integration.
∫ dx + 2 ∫ dx = x + 2x + C2 = 3x + C2, where C2 is another constant of integration.
Therefore, the equation becomes:
ln|y + 5| + C1 = 3x + C2
To simplify further, you can combine the constants of integration into a single constant C:
ln|y + 5| = 3x + C
Exponentiating both sides, you get:
|y + 5| = e^(3x + C)
Since e^C is just a constant, you can write it as K:
|y + 5| = Ke^(3x)
Now, we need to consider two cases based on the absolute value:
1. y + 5 > 0:
If y + 5 > 0, then |y + 5| = y + 5, and the equation becomes:
y + 5 = Ke^(3x)
2. y + 5 < 0:
If y + 5 < 0, then |y + 5| = -(y + 5), and the equation becomes:
-(y + 5) = Ke^(3x)
Now, you can solve these two cases separately.
Case 1: y + 5 > 0
Solving for y:
y + 5 = Ke^(3x)
y = Ke^(3x) - 5
Case 2: y + 5 < 0
Solving for y:
-(y + 5) = Ke^(3x)
y = -Ke^(3x) - 5
So, the solution to the initial value problem dy/dx = (y + 5)(x + 2) is given by:
1. If y + 5 > 0, then y = Ke^(3x) - 5
2. If y + 5 < 0, then y = -Ke^(3x) - 5
You can substitute the initial value to determine the constant of integration K and get the final solution.