Question

If the differential equation having $y = Ae^x + B\sin x$ as its general solution is $f(x)\frac{d^2y}{dx^2}+g(x)\frac{dy}{dx}+h(x)y=0$, then $f(x)+g(x)+h(x) =$

• A. $2\cos x$
• B. $4\sin x$
• C. 0
• D. $\cos x - \sin x$

Hint:

To find the differential equation from a general solution with $n$ arbitrary constants, differentiate the solution $n$ times. This creates a system of $n+1$ equations. Eliminate the constants from this system to obtain the differential equation.

Answer 1

The Correct Option is C

The general solution is $y = Ae^x + B\sin x$. To find the differential equation, we need to eliminate the arbitrary constants A and B. This requires differentiating twice.
First derivative:
$y' = \frac{d}{dx}(Ae^x + B\sin x) = Ae^x + B\cos x$.
Second derivative:
$y'' = \frac{d}{dx}(Ae^x + B\cos x) = Ae^x - B\sin x$.
We have a system of three equations:
(1) $y = Ae^x + B\sin x$
(2) $y' = Ae^x + B\cos x$
(3) $y'' = Ae^x - B\sin x$
From (1) and (3), we can write: $y - y'' = (Ae^x + B\sin x) - (Ae^x - B\sin x) = 2B\sin x$.
So, $B\sin x = \frac{y-y''}{2}$.
From (1) and (3), we can also write: $y+y'' = (Ae^x + B\sin x) + (Ae^x - B\sin x) = 2Ae^x$.
So, $Ae^x = \frac{y+y''}{2}$.
Now substitute these expressions for $Ae^x$ and $B\sin x$ into equation (2). First we need an expression for $B\cos x$.
From $B\sin x = \frac{y-y''}{2}$, we get $B = \frac{y-y''}{2\sin x}$. Then $B\cos x = \frac{(y-y'')\cos x}{2\sin x}$.
Substitute into (2): $y' = \frac{y+y''}{2} + \frac{(y-y'')\cos x}{2\sin x}$.
Multiply by $2\sin x$: $2y'\sin x = (y+y'')\sin x + (y-y'')\cos x$.
$2y'\sin x = y\sin x + y''\sin x + y\cos x - y''\cos x$.
Rearrange to the form $f(x)y''+g(x)y'+h(x)y=0$:
$y''(\sin x - \cos x) - y'(2\sin x) + y(\sin x + \cos x) = 0$.
Or, multiplying by -1: $y''(\cos x - \sin x) + y'(2\sin x) - y(\sin x + \cos x) = 0$. Wait, sign error. Backtracking: $y''(\cos x - \sin x) + y'(2\sin x) + y(-\sin x - \cos x) = 0$.
Comparing with the standard form, we have:
$f(x) = \cos x - \sin x$
$g(x) = 2\sin x$
$h(x) = -\sin x - \cos x$
We need to find the sum $f(x)+g(x)+h(x)$:
$(\cos x - \sin x) + (2\sin x) + (-\sin x - \cos x) = \cos x - \sin x + 2\sin x - \sin x - \cos x = 0$.