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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Solve the differential equation $\begin{bmatrix}\large\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\large\frac{y}{\sqrt{x}}\end{bmatrix}\large\frac{dx}{dy}$$=1(x\neq0)$

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  • To solve the first order linear differential equation:
  • (i) First write the equation of the form $\large\frac{dy}{dx}$$ + py = Q$
  • (ii) Find the integral factor(I.F) =$ e^{\int pdx}$
  • (iii) Write the solution as y.(I.F) - integration of Q.(I.F) + C
Step 1:
Given : $\big[\large\frac{e^{-2\sqrt x}}{\sqrt x}-\large\frac{y}{\sqrt x}\big]\large\frac{dx}{dy}$$=1$
$\large\frac{dx}{dy}=\frac{1}{\big[\Large\frac{e^{-2\sqrt x}}{\sqrt x}-\frac{y}{\sqrt x}\big]}$
$\large\frac{dy}{dx}=\big[\Large\frac{e^{-2\sqrt x}}{\sqrt x}-\frac{y}{\sqrt x}\big]$
$\large\frac{dy}{dx}+\frac{y}{\sqrt x}=\frac{e^{-2\sqrt x}}{\sqrt x}$
Here $p=\large\frac{1}{\sqrt x}$
$Q=\Large\frac{e^{-2\sqrt x}}{\sqrt x}$
Step 2:
To find the integrating factor (I.F) integrate p.
$\int \large\frac{1.dx}{\sqrt x}=$$2\sqrt x$
Hence $I.F=e^{2\sqrt x}$
Step 3:
Hence the solution is $y.e^{2\sqrt x}$=$\int \large\frac{e^{-2\sqrt x} }{\sqrt x}.$$e^{2\sqrt x}dx$
$y.e^{\sqrt x}=\int \large\frac{dx}{\sqrt x}$
$y.e^{\sqrt x}=2\sqrt x+C$
Hence this is the required solution.
answered Jul 30, 2013 by sreemathi.v

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