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Find the general solution of the differential equation $x\large\frac{dy}{dx}$$+2y=x^2\log x$

1 Answer

  • 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})$$.dx$
$\qquad\qquad\qquad\qquad\qquad\quad=\log x.(\large\frac{x^4}{4}) - (\large\frac{1}{4})\frac{x^4}{4}$
Hence the required solution is $yx^2 = \log x(\large\frac{x^4}{4}) -\frac{ x^4}{16}$
$yx^2 = (\large\frac{1}{16})$$x^{\large[4\log x - 1]}$
This is the required solution.
answered Aug 1, 2013 by sreemathi.v
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