InterviewSolution
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InterviewSolution
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Find the general solution of the differential equation `(dy)/(dx)-y=cosx` |
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Answer» The given differential equation is ` (dy)/(dx) -y = cos x " " ` ... (i) This is of the form ` (dy)/(dx) + P y = Q, ` where ` P =- 1 and Q = cos x ` Thus, the given differential equation is linear. `IF= e ^(int Pdx)= e ^( int - dx) = e ^(-x )` So, the required solution is `y xx IF = int { Q xxIF} dx + C, ` i.e., `y xx e ^(-x) = int (cos x ) e ^(-x) dx + C " "` ... (ii) Let `I =int (cos x ) e ^(-x) dx ` ` =( cos x ) (- e ^(-x)) - int (-sin x ) (-e ^(- x )) dx ` [ integrating by parts] `" " = - (cos x ) e ^(- x) - int underset ("I")((sinx )) underset("II")((e^(-x)))dx` ` " " = - (cos x ) e ^(-x) - { (sin x ) (-e^(-x)) } - int (cos x ) (-e^(-x)) dx ` ` = - ( cos x ) e ^(-x) + (sin x ) (e ^(-x)) -I.` `therefore 2 I = (sin x - cos x ) e^(-x) rArr I = (1)/(2) (sinx - cos x ) e ^(-x)`. Putting the value in (ii), we get ` y xx e ^(-x) = (1)/(2) ( sinx - cos x ) e ^(-x) + C rArr y = (1)/(2)(sin x - cos x ) + Ce^(x)` Hence, `y = (1)/(2)( sinx - cos x ) + Ce ^(x) ` is the required solution. |
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