# Find the general solution of the differential equation $x\large\frac{dy}{dx}$$+2y=x^2\log x ## 1 Answer Toolbox: • To solve the first order linear differential equation of the form \large\frac{dy}{dx}$$ + Py = Q$
• (i) Write the given equation in the form of $\large\frac{dy}{dx}$$+ Py = Q • (ii) Find the integrating factor (I.F) = e^{\int Pdx}. • (iii) Write the solution as y(I.F) = integration of Q(I.F) dx + C Step 1: The given problem is a first order linear equation. To write it in the form, let us divide throughout by x \large\frac{dy}{dx} +\frac{2y}{x} =$$ x \log x$
The I.F = $\int \large\frac{2}{x }$$= 2\log x or \log x^2 The I.F = e^{\log x^2} = x^2 The solution is yx^2 = \int x^2\log x . x^2 + C yx^2 =\int x^3\log x.dx + C \int x^3 \log x dx can be done by integration by parts. Step 2: Let u = \log x;du = \large\frac{dx}{x} and dv = x^3dx or v =\large\frac{x^4}{4} Hence integration of x^3\log xdx = \log x(\large\frac{x^4}{4}) - \int \large\frac{1}{x}(\large\frac{x^4}{4})$$.dx$
$\qquad\qquad\qquad\qquad\qquad\quad=\log x.(\large\frac{x^4}{4}) - (\large\frac{1}{4})\frac{x^4}{4}$
Hence the required solution is $yx^2 = \log x(\large\frac{x^4}{4}) -\frac{ x^4}{16}$