Q. 19

# Find the particul

yey dx = (y3 + 2xey) 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 Rate this question :

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