Question

Dy/dx+(2xtan-¹y-x³)(1+y²)=0 semister 2nd bsc

Answer 1

$\implies \sf \: \dfrac{dy}{dx} + [2x + (tan^ {- 1}y)- (x3) ][1+y^2]$

Now, Divide the equation by :

$\implies \sf \: 1+y ^{2} \dfrac{dy}{dx}( \dfrac{1}{1} +y^{2} ) + [2x + tan^{ - 1}y - x3 ]$

_______________[ Equation 1 ]

Now, Putting :

$\implies \sf \: tan{^-1} \: y=t$

Differentiate wrt x,

$\implies \sf \dfrac{dy}{dx}( \dfrac{1}{1}+y^{2} ) = \dfrac{dt}{dt}$

_______________[ Equation 2 ]

Now, Substitute the values of Equation 1 in Equation 2.

So,

$\implies \sf \dfrac{dt}{dx} + t +2x-x ^{3} = 0$

Now, It is linear differential equation.

It's integral factor is ex.

____________________

Answer 2

Given,

$\frac{dy}{dx}+(2x\tan^{-1}y-x^3)(1+y^2)$

Dividing by (1 + y²)

$\frac{1}{1+y^2}\frac{dy}{dx}+(2x\tan^{-1}y-x^3)=0\\\;\\\text{Putting}\;\tan^{ - 1}y=t\\\;\\\implies\frac{1}{1+y^2}\frac{dy}{dx}=\frac{dt}{dx}\\\;\\\textbf{On doing above substitutions,}\\\;\\\frac{dt}{dx}+2xt-x^3 = 0\\\;\\\frac{dt}{dx}+2xt=x^3$

This is the Linear Diffrentiatial Equation in t.

$\text{I.F.}=e^{\int{2x\,dx}}\\\;\\\text{I.F.}=e^{x^2}$

Thus solution of this diffrentiatial equation will be,

$t\times\text{I.F.}=\int{x^2\times \text{I.F.}\,dx}+c\\\;\\t\times e^{x^2}=\int{x^3e{x^2}\,dx}+c\\\;\\te^{x^2}=e^{x^2}(x^2-1)+c\\\;\\\textbf{Putting the value of t,}\\\;\\e^{x^2}\tan^{-1}y=e^{x^2}(x^2-1)+c$

This is the required General Solution of given Diffrentiatial Equation.