Math  /  Calculus

QuestionQ4) Find the differential equation whose general solution y=Acos(lnx)+Bsin(lnx),x>0y=A \cos (\ln x)+B \sin (\ln x), x>0.

Studdy Solution
Eliminate the arbitrary constants A A and B B to form the differential equation.
Notice that:
y=Acos(lnx)+Bsin(lnx) y = A \cos (\ln x) + B \sin (\ln x)
dydx=Asin(lnx)x+Bcos(lnx)x \frac{dy}{dx} = -\frac{A \sin (\ln x)}{x} + \frac{B \cos (\ln x)}{x}
d2ydx2=A(cos(lnx)sin(lnx))x2+B(sin(lnx)cos(lnx))x2 \frac{d^2y}{dx^2} = -\frac{A (\cos (\ln x) - \sin (\ln x))}{x^2} + \frac{B (-\sin (\ln x) - \cos (\ln x))}{x^2}
To eliminate A A and B B , observe that the coefficients of cos(lnx)\cos (\ln x) and sin(lnx)\sin (\ln x) in the derivatives can be combined to form a differential equation.
The differential equation is:
x2d2ydx2+xdydx+y=0 x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = 0
The differential equation whose general solution is y=Acos(lnx)+Bsin(lnx) y = A \cos (\ln x) + B \sin (\ln x) is:
x2d2ydx2+xdydx+y=0 \boxed{x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = 0}

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