If y′= x(1+ y) and y > -1 , then y =
The differential equation dy/dx=y/x^2 has a solution given by:
dy/y = dx/x^2
ln y = -1/x + c
y = C e^-1/x
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http://www.wolframalpha.com/input/?i=Solve++y%27(x)+%3D+x%2Bxy
For the first question, let's first separate the variables and integrate:
dy/(1+y) = x dx
Integrating both sides gives us:
ln|1+y| = (1/2)x^2 + C1
Now, let's solve for y by using properties of natural logarithms:
|1+y| = e^((1/2)x^2 + C1)
Since y > -1, we can eliminate the absolute value:
1+y = e^((1/2)x^2 + C1)
y = e^((1/2)x^2 + C1) - 1
And that's the solution for y.
For the second question, let's first rewrite the differential equation as:
y' = y/x^2
Now, this is a separable differential equation. Let's separate the variables and integrate:
dy/y = dx/x^2
Integrating both sides gives us:
ln|y| = -1/x + C2
Now, let's solve for y by using properties of natural logarithms:
|y| = e^(-1/x + C2)
Since y can be positive or negative, we can eliminate the absolute value:
y = ± e^(-1/x + C2)
And that's the solution for y.
To solve the differential equation y' = x(1 + y), we can use separation of variables.
1. Start by separating the variables by moving the y term to one side and the x term to the other side:
dy/(1 + y) = x dx
2. Integrate both sides with respect to their respective variables:
∫(1 + y)^(-1) dy = ∫x dx
To integrate the left side, we can use the substitution method. Let u = 1 + y, then du = dy:
∫u^(-1) du = ∫x dx
ln|u| = 1/2 x^2 + C1, where C1 is the constant of integration.
3. Substitute back in for u:
ln|1 + y| = 1/2 x^2 + C1
4. Now, we can solve for y. Start by exponentiating both sides:
e^(ln|1 + y|) = e^(1/2 x^2 + C1)
|1 + y| = e^(1/2 x^2 + C1)
5. Since y > -1, we can remove the absolute value:
1 + y = e^(1/2 x^2 + C1)
y = e^(1/2 x^2 + C1) - 1
This is the general solution to the differential equation y' = x(1 + y) with the condition y > -1.
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For the second question about the differential equation dy/dx = y/x^2, we will also use separation of variables.
1. Start by separating the variables:
dy/y = dx/x^2
2. Integrate both sides:
∫(1/y) dy = ∫(1/x^2) dx
To integrate the left side, we can use the natural logarithm:
ln|y| = -1/x + C2, where C2 is the constant of integration.
3. Exponentiate both sides:
|y| = e^(-1/x + C2)
4. Since y can be positive or negative, we can write it as a absolute value:
|y| = e^(-1/x + C2)
5. Now we have two cases:
Case 1: y > 0
For y > 0, we can remove the absolute value:
y = e^(-1/x + C2)
Case 2: y < 0
For y < 0, we take the negative sign outside the absolute value:
-y = e^(-1/x + C2)
y = -e^(-1/x + C2)
So, the general solution to the differential equation dy/dx = y/x^2 includes two cases: y = e^(-1/x + C2) for y > 0 and y = -e^(-1/x + C2) for y < 0.
more like
dy/(1+y) = x dx
ln(1+y) = x^2/2
1+y = c e^(x^2/2)
y = c e^(x^2/2) - 1