The order of the differential equation whose | vedclass

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(C) The given general solution is $y = (c_1 + c_2) \cos(x + c_3) - c_4 e^{x + c_5}$.Let $A = (c_1 + c_2)$ and $B = c_4 e^{c_5}$. Then the equation sim

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$3$

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(C) The given general solution is $y = (c_1 + c_2) \cos(x + c_3) - c_4 e^{x + c_5}$.Let $A = (c_1 + c_2)$ and $B = c_4 e^{c_5}$. Then the equation simplifies to $y = A \cos(x + c_3) - B e^x$.Here,we have three independent arbitrary constants: $A$,$c_3$,and $B$.To find the order of the differential equation,we differentiate $y$ with respect to $x$ three times:$y' = -A \sin(x + c_3) - B e^x$$y'' = -A \cos(x + c_3) - B e^x$$y''' = A \sin(x + c_3) - B e^x$From the original equation,$A \cos(x + c_3) = y + B e^x$. Substituting this into $y''$,we get $y'' = -(y + B e^x) - B e^x = -y - 2B e^x$.Thus,$2B e^x = -y - y''$.Differentiating this again,$2B e^x = -y' - y'''$.Equating the two expressions for $2B e^x$,we get $-y - y'' = -y' - y'''$,which simplifies to $y''' - y'' + y' + y = 0$.Since the highest derivative is of order $3$,the order of the differential equation is $3$.

The order of the differential equation whose general solution is given by $y = (c_1 + c_2) \cos(x + c_3) - c_4 e^{x + c_5}$,where $c_1, c_2, c_3, c_4, c_5$ are arbitrary constants,is

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