# Solve the following differential equation : $(1+x^2)\large\frac{dy}{dx}$$-2xy=(x^2+2)(x^2+1) ## 1 Answer Toolbox: • To solve the first order linear differential equation of the form \large\frac{dy}{dx}$$ + Py = Q$
• (i) Write the given equation in the form of $\large\frac{dy}{dx}$$+ Py = Q • (ii) Find the integrating factor (I.F) = e^{\int Pdx}. • (iii) Write the solution as y(I.F) = integration of Q(I.F) dx + C Step 1: (1+x^2)\large\frac{dy}{dx}$$-2xy=(x^2+2)(x^2+1)$
Clearly this is a linear differential equation .
Divide throughout by $(1+x^2)$
$\large\frac{dy}{dx}-\frac{2x}{1+x^2}$$y=x^2+2 This is of the form \large\frac{dy}{dx}$$+Py=Q$
Hence $P=\large\frac{-2x}{1+x^2}$ and $Q=x^2+2$
The general solution for this equation is
$ye^{\int Pdx}=\int Qe^{\int Pdx}dx+c$
Step 2:
Let us find the integral factor I.F
$\int Pdx=\int \large\frac{2x}{1+x^2}$$dx Put 1+x^2=t,then 2xdx=dt \therefore \int\large\frac{dt}{t}$$=\log t$
$\qquad\;\;=\log (1+x^2)$
$\therefore e^{\int Pdx}=e^{\log (1+x^2)}$
$\Rightarrow 1+x^2$
$\therefore y(1+x^2)=\int (x^2+2).(1+x^2)dx$
Step 3:
$y(1+x^2)=\int [x^4+2x^2+x^2+2]$
$y(1+x^2)=\int( x^4+3x^2+2)dx$
On integrating we get,
$y(1+x^2)=\large\frac{x^5}{5}+\frac{3x^3}{3}$$+2x+c y(1+x^2)=\large\frac{x^5}{5}$$+x^3+2x+c$
$5y(1+x^2)=x^5+5x^3+10x+c$