Question

Find the general solution: $x\frac {dy}{dx}+2y=x^2log\ x$

Answer 1

The given differential equation is:

x$\frac {dy}{dx}$+2y = x2logx

⇒$\frac {dy}{dx}$+$\frac {2}{xy}$ = xlog x

This equation is in the form of a linear differential equation as:

$\frac {dy}{dx}$+py = Q(where p=$\frac 2x$ and Q=xlog x)
∫pdx = 2∫$\frac 1x$ dx = 2log x

Now, I.F. = e∫pdx = e2logx = $e^{log x^2}$ = x2

The general solution of the given differential equation is given by the relation,

y(I.F.) = ∫(Q×I.F.)dx + C

⇒y.x2 = ∫(xlog x . x2)dx + C

⇒x2y = ∫(x3log x)dx + C

⇒x2y = logx.∫x3dx - ∫[$\frac {d}{dx}$(logx).∫x3dx]dx + C

⇒x2y = log x . $\frac {x^4}{4}$-∫($\frac {1}{x}$.$\frac {x^4}{4}$)dx + C

⇒x2y = x4log $\frac {x^4}{4}$-$\frac {1}{4}$∫x3dx + C

⇒x2y = x4log$\frac {x^4}{4}$-$\frac {1}{4}$. $\frac {x^4}{4}$ + C

⇒x2y = $\frac {1}{16}$x4(4logx-1) + C

⇒y = $\frac {1}{16}$x2(4log x-1) + Cx-2