Q. 19

Find the particul

Answer :

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



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