Question

The order of the differential equation whose general solution is given by $y=c_1{e}^{2x+C_2}+C_3{e}^x+C_4\sin (x+C_5)$ is?

• A. $5$
• B. $4$
• C. $3$
• D. $2$

Answer 1

The correct option is B $4$

We are given $y=C_1{e}^{2a+C_2}+C_3{e}^x+X_4.\sin (x+C_5)$

as we know that

$e^{a+b}=e^a.e^b$ & $\sin (A+B)=\sin A\cos B+\cos A\sin B$

So, $y=C_1.e^{2x}.e^{C_2}+C_3{e}^x+C_4$.

$\left[\right(\sin x\cos C_5+\cos x\sin C_5\left)\right]$

$y=(C_1.e^{C_2})e^{2x}+C_3{e}^x+(C_4.\cos C_5)\sin x+(C_4.\sin C_5)\cos x$

here $e^{C_2},\cos C_5$ & $\sin C_5$ are constant

terms So, let $C_1=C_1.e^{C_2}$

& $C_4=\cos C_5.C_4$ & $C_5=C_4.\sin C_5$

So, $y=C_1{e}^{2x}+C_3{e}^x+C_4\sin x+C_5\cos x$

So as we know the order of the differential equation are defined as the no of applicable arbitary constant as $C_2,C_3,C_4$ & $C_5$ are $4$ arbitary constant So the order is $4$

Answer is $4$