Find the particular solution of the differential equation:

ye^y dx=(y³+2xe^y) dy, y (0) =1

OR

Show that (x - y) dy = (x + 2y)dx is a homogenous differential equation. Also, find the general solution of the given differential equation.

ye^y dx = (y³ + 2xe^y) dy

Compare with

This is linear differential equation in x

For the solution of linear differential equation, we first need to find the integrating factor

⇒ IF = e^∫Pdy

The solution of linear differential equation is given by x(IF) = ∫Q(IF)dy + c

Substituting values for Q and IF

Now we have to find c for that we have given y(0) = 1 which means when x = 0 we have y = 1

Putting these values

⇒ 0 = - e^-1 + c

⇒ c = e^-1

Put the value of c and the solution of differential equation would be

⇒ x = (e^-1 – e^-y) y²

The solution of given differential equation is x = (e^-1 – e^-y)y²

OR

f(x, y) is said to be homogenous equation if f(λx, λy) = λⁿf(x, y) where λ ≠ 0

Now, a differential equation is said to be homogenous if f(x, y) is homogenous

Given equation

(x - y)dy = (x + 2y)dx

Let us check whether f (x, y) is homogenous or not

Put λx, λy in place of x, y respectively

⇒ f (λx, λy) = λ⁰f(x, y)

Hence by definition f (x, y) is homogenous

As f (x, y) is homogenous is homogenous

Hence is a homogenous equation.

Now we have to solve that equation

Put y = vx

Using uv rule

Multiply divide by -2 in LHS

Now integrate

Observe that is of the form whose solution is log[f(v)].

Note: can be proved by substituting f(v) = t hence dt = f’(v)dv which gives

Hence and

Hence equation (i) becomes

Now to solve the integral we have to somehow make perfect square in the denominator

As we know that (as integration is inverse of differentiation)

Resubstitute

We added the integration constant in the last step because it is the combined one for all the integrals

Hence solution of differential equation

As log x + log y = log xy