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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Solve the following differential equation : $\large \frac{dy}{dx}$$+y=\cos\: x-\sin\: x$

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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 $Q(I.F) dx + C$
  • $\int e^x[f(x)+f'(x)]dx=e^xf(x)+c$
Step 1:
Given :$\large\frac{dy}{dx}$$+y=\cos x-\sin x$
This is a linear differential equation of the form
$\large\frac{dy}{dx}$$+Py=Q$
Where $P=1$ and $Q=\cos x-\sin x$
$\int Pdx=\int dx$
$\qquad\;\;=x$
Step 2:
Now the solution for the above equation :
$ye^\int Pdx=\int Qe^{\int P dx}dx+c$
$ye^x=\int e^x(\cos x-\sin x)+c$
Step 3:
But $\int e^x[f(x)+f'(x)]dx=e^xf(x)+c$
$\therefore \int e^x(\cos x-\sin x)=e^x\cos x+c$
Hence the required solution is
$ye^x=e^x\cos x+c$
Dividing throughout by $e^x$ we get,
$\Rightarrow y=\cos x+ce^{-x}$
answered Sep 23, 2013 by sreemathi.v
 
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