Solve xy' + y + y^2 = 0 for y(1) = 2
---------
I isolated y' such that: y' = (-y^2-y)/x
But I'm not sure what to substitute from here
I could do v = -y^2- y and v' = -2yy' - y' but I'm not sure this would help
just for fun
x dy/dx = -y^2 - y
- dy /[y(y+1) ] = dx/x
-1 ln [ y/(y+1) ] = ln x + ln c = ln xc
ln [ y/(y+1) ]^-1 = ln xc
x c = (y+1)/y
if x = 1 , y = 2
c = 3/2
3 x = 2(y+1)/y
I'm not sure if your integration is correct; I got -ln|1/y + 1|
However, thank you. Is substitution even possible for this problem?
I used my 1955 integration tables
int { dx/[x(ax+b)] } = (1/b)ln [ x/(ax+b) ] + constant
I did it that way because I did not figure out how o do it with substitution.
use partial fractions.
1/(y(y+1)) = 1/y - 1/(y+1)
To solve the given differential equation xy' + y + y^2 = 0 with the initial condition y(1) = 2, you can use a technique called separation of variables.
First, let's rewrite the equation in a more standard form by dividing through by x:
y' + (y + y^2)/x = 0
Now, separate the variables by multiplying both sides by dx:
dy + (y + y^2)/x dx = 0
Next, let's separate the terms involving y and the terms involving x:
dy + y/x dx + y^2/x dx = 0
Now, let's focus on the terms involving y:
dy + (y/x + y^2/x) dx = 0
To integrate both sides, we can rewrite the equation as:
dy + y(1/x + y/x) dx = 0
Now, let's define a new variable u = 1/x, which means du/dx = -1/x^2. Rewriting the equation in terms of u:
dy + y(u + y) du = 0
This equation is separable. Now, we can integrate both sides:
∫ (1/y) dy + ∫ (u + y) du = ∫ 0 dx
ln|y| + (u/2 + y^2/2) + C = 0
Here, C is an arbitrary constant that arises from integration. Let's simplify the equation:
ln|y| + (1/(2x) + y^2/2) + C = 0
Now, apply the initial condition y(1) = 2 to find the value of C:
ln|2| + (1/(2*1) + 2^2/2) + C = 0
ln|2| + (1/2 + 2) + C = 0
ln|2| + 5/2 + C = 0
C = -ln|2| - 5/2
Therefore, the final solution to the differential equation is:
ln|y| + (1/(2x) + y^2/2) - ln|2| - 5/2 = 0