Browse Questions

# Solve the differential equation $(1+y^2)\tan^{-1}xdx+2y(1+x^2)dy=0$

Toolbox:
• If a linear differential equation is of the form $\large\frac{dy}{dx}$$=f(x), then it can be solved by seperating the variables. • \int \large\frac{dx}{x^2+a^2}=\frac{1}{a}$$\tan ^{-1} \bigg(\large\frac{x}{a}\bigg)+c$
Given $(1+y^2)\tan^{-1}xdx+2y(1+x^2)dy=0$
We can rewrite this as
$2y(1+x^2)dy=-(1+y^2)\tan ^{-1} x dx$
$2 \large\frac{dy}{dx}=\frac{-(1+y^2)\tan ^{-1}x}{(1+x^2)}$
Now seperating the variables we get
$2 \large\frac{dy}{1+y^2}=\frac{-\tan ^{-1}(x)}{1+x^2}$$dx Now integrating we get, \int 2 \large\frac{dy}{1+y^2}=-\int\frac{\tan ^{-1}(x)}{1+x^2}$$dx$
Consider $\large\frac{dy}{1+y^2}$
This is of the form $\int \large\frac{dx}{x^2+a^2}=\frac{1}{a} $$\tan ^{-1} \bigg(\large\frac{x}{a}\bigg) Hence 2 \int \large\frac{dy}{1+y^2}$$=2 \tan ^{-1}(y)$
Step 2:
Consider $\int \large\frac{-\tan ^{-1}(x)}{1+x^2}$$dx Let \tan ^{-1}x=t on differentiating w.r.t x we get \large\frac{1}{1+x^2}$$dx=dt$
Now substituting this we get,
$\large\frac{-1}{2} $$\int t.dt On integrating we get -\bigg[\large\frac{t^2}{2}\bigg] substituting for t we get =\large\frac{-(\tan ^{-1}x)^2}{2} Now applying this in required solution we get, 2\tan ^{-1}y=\large\frac{-(\tan ^{-1}x)^2}{2}+c 2\tan ^{-1}y+\large\frac{(\tan ^{-1}x)^2}{2}=c Hence the required solution is 2\tan ^{-1}y+\large\frac{1}{2}$$(\tan ^{-1}x)^2=c$