Solve the following differential equation: (1 + x)*(dy/dx) = y

where y is a funtion of X

solve by the following:

A)series expansion
B) elemantary method

please show working

A) Series Expansion Method:

To solve the differential equation (1 + x)*(dy/dx) = y using series expansion, we can assume that y can be represented as a power series:

y = a0 + a1x + a2x^2 + a3x^3 + ...

where a0, a1, a2, a3, ... are the coefficients to be determined.

Now, let's find the derivatives of y with respect to x:

dy/dx = a1 + 2a2x + 3a3x^2 + 4a4x^3 + ...

Substituting these into the given differential equation, we have:

(1 + x)*(a1 + 2a2x + 3a3x^2 + 4a4x^3 + ...) = (a0 + a1x + a2x^2 + a3x^3 + ...)

Expanding the left-hand side, we get:

a1 + (2a2 + a1)x + (3a3 + 2a2)x^2 + (4a4 + 3a3)x^3 + ...

We can equate the coefficients of corresponding powers of x on both sides of the equation to obtain a system of equations.

For example, when equating the coefficients of x^0, we have:

a1 = a0

When equating the coefficients of x^1, we have:

2a2 + a1 = a1

Simplifying this, we get:

2a2 = 0

We can continue this process to find the values of other coefficients. The general equation for the nth coefficient is given by:

(n+1)an + nan-1 = an-1

Simplifying this equation, we can express the coefficient an in terms of the previous one an-1:

an = an-1 / (n+1)

Starting with a0 = a1, we can use this recursive formula to determine all the coefficients.

Once the coefficients are determined, we can substitute them back into the power series representation of y to obtain the solution.

B) Elementary Method:

The given differential equation (1 + x)*(dy/dx) = y is a first-order linear homogeneous ordinary differential equation.

Let's rewrite the equation as:

(1/x) dx = (1/y) dy

Now, let's integrate both sides:

∫(1/x) dx = ∫(1/y) dy

ln|x| + C1 = ln|y| + C2

where C1 and C2 are integration constants.

Combining the constants:

ln|x| = ln|y| + C

where C = C2 - C1 is a new integration constant.

Taking the exponential of both sides:

|x| = e^(ln|y| + C)

|x| = e^(ln|y|) * e^C

|x| = |y| * K

where K = e^C is a constant.

We can rewrite this equation as:

y = ± (K * |x|)

Considering the plus and minus signs, we have two possible solutions:

1) y = K * x
2) y = -K * x

Both of these equations satisfy the original differential equation.

Thus, the general solution to the differential equation is:

y = K * x, or y = -K * x, where K is an arbitrary constant.