If y′= x(1+ y) and y > -1 , then y =
The differential equation dy/dx=y/x^2 has a solution given by:
To find the solution to the differential equation y′= x(1+ y), we can rearrange the equation and separate the variables.
First, divide both sides of the equation by (1+y):
dy / (1+y) = x dx
Now, we can integrate both sides:
∫(1/(1+y)) dy = ∫x dx
To integrate the left side, we can use the substitution u = 1+y and du = dy:
∫(1/u) du = ∫x dx
ln|u| = (1/2)x^2 + C (where C is the constant of integration)
Now, substitute back u = 1+y:
ln|1+y| = (1/2)x^2 + C
To solve for y, we can take the exponential of both sides:
e^(ln|1+y|) = e^((1/2)x^2 + C)
|1+y| = e^((1/2)x^2 + C)
Since y > -1, we can remove the absolute value:
1+y = e^((1/2)x^2 + C)
Now, rearrange the equation to solve for y:
y = e^((1/2)x^2 + C) - 1
This is the solution to the differential equation y′= x(1+ y), where y > -1.
----------------------------------------------------------
To find the solution to the differential equation dy/dx = y/x^2, we can separate the variables and integrate.
First, rearrange the equation:
dy = (y/x^2) dx
Now, separate the variables:
(1/y) dy = (1/x^2) dx
Next, integrate both sides:
∫(1/y) dy = ∫(1/x^2) dx
To integrate the left side, we can use the substitution u = y and du = dy:
∫(1/u) du = ∫(1/x^2) dx
ln|u| = -1/x + C (where C is the constant of integration)
Now, substitute back u = y:
ln|y| = -1/x + C
To solve for y, we can take the exponential of both sides:
e^(ln|y|) = e^(-1/x + C)
|y| = e^(-1/x + C)
Since y can be positive or negative, we can remove the absolute value:
y = ± e^(-1/x + C)
This is the solution to the differential equation dy/dx = y/x^2.