(1+x)dy-ydx=0

4 min read Jun 16, 2024

Solving the Differential Equation (1+x)dy - ydx = 0

The given differential equation (1+x)dy - ydx = 0 is a first-order homogeneous differential equation. This means it can be written in the form:

dy/dx = f(y/x)

Let's solve this equation step by step:

1. Rearrange the equation:

First, we rearrange the equation to get it in the standard form:

(1+x)dy = ydx

dy/dx = y/(1+x)

2. Substitution:

We introduce a new variable, v = y/x. This implies y = vx and dy = vdx + xdv.

Substituting these into our rearranged equation, we get:

(vdx + xdv)/dx = vx/(1+x)

3. Simplify and Solve:

Simplifying the equation, we get:

v + x(dv/dx) = vx/(1+x)

Rearranging to separate variables:

x(dv/dx) = vx/(1+x) - v

x(dv/dx) = v(x - 1 - x)/(1+x)

x(dv/dx) = -v/(1+x)

dv/v = -dx/x(1+x)

Now we have separated the variables. We can integrate both sides:

∫(dv/v) = -∫(dx/x(1+x))

The integral on the left side is ln|v|. To solve the integral on the right side, we use partial fractions:

1/(x(1+x)) = A/x + B/(1+x)

Solving for A and B, we get A = 1 and B = -1. Therefore:

-∫(dx/x(1+x)) = -∫(dx/x) + ∫(dx/(1+x)) = -ln|x| + ln|1+x|

4. General Solution:

Combining the integrals, we have:

ln|v| = -ln|x| + ln|1+x| + C

where C is the constant of integration. Using the properties of logarithms, we can simplify:

ln|v| = ln|(1+x)/x| + C

ln|v| - ln|(1+x)/x| = C

ln|vx/(1+x)| = C

|vx/(1+x)| = e^C

Since e^C is a positive constant, we can write it as K:

|vx/(1+x)| = K

vx/(1+x) = ±K

Finally, substituting back v = y/x:

(y/x)x/(1+x) = ±K

y/(1+x) = ±K

y = ±K(1+x)

This is the general solution of the given differential equation.

Conclusion

We have successfully solved the first-order homogeneous differential equation (1+x)dy - ydx = 0. The solution is a family of curves represented by y = ±K(1+x), where K is an arbitrary constant. This solution can be used to describe various physical phenomena that can be modeled by this differential equation.