Browse Questions

# Find the equation of a curve passing through origin and satisfying the differential equation $(1-x^2)\Large \frac{dy}{dx}\normalsize-2xy=4x^2$

Toolbox:
• If the given linear differential equation is of the form $\large\frac{dy}{dx}$$+Py=Q, then its general solution is ye^{\int pdx}=\int Q e ^{\int pdx}.dx+c • \int\large\frac{dx}{x+a}$$=\log |x+a|+c$
Step 1:
Given equation of the curve $(1-x^2)\large\frac{dy}{dx}$$-2xy=4x^2 Divide throughout by (1-x^2) \large\frac{dy}{dx}-\frac{2x}{1-x^2}.$$y=\large\frac{4x^2}{1-x^2}$
Clearly this is a linear differential equation of the form
$\large\frac{dy}{dx}$$+P.y=Q Where P=-\large\frac{2x}{1-x^2} and Q=\large\frac{4x^2}{1-x^2} Let us now find the integrating factor I.F e^{\int pdx}=e^{\int \Large\frac{-2x}{1-x^2}}dx Step 2: Consider \int \large\frac{-2x}{1-x^2}$$dx$
Put $1-x^2=t$ on differentiating w.r.t t
We get $-2xdx=dt$
Now substituting this we get,
$\large\int \frac{dt}{t}$$=\log |t| Substituting for t we get, \log |1-x^2| Hence I.F is e^{\large\log(1-x^2)} but e^{\large\log x}$$=x$
Hence $I.F=1-x^2$
Now the required solution is
$ye^{\large\int pdx}$$=\int Q. e ^{\large\int pdx}$$dx+c$
$y.(1-x^2)=\int \large\frac{4x^2}{1-x^2}$$(1-x^2)dx+c$
=>$y(1-x^2)=\int 4x^2.dx+c$
On integrating we get
$y(1-x^2)=\large\frac{4x^3}{3}+c$
Step 3:
It is given that the curve passes through the orgin, which means $x=0,y=0$
Let us now substitute for x and y to evaluate the value for c
$0(1+0)=4 \times 0+c=>c=0$
Hence the required solution is
$y(1-x^2)=\large\frac{4x^3}{3}$=>$3y(1-x^2)=4x^3$
=>$y=\large\frac{4x^3}{3(1-x^2)}$