Find the differential equation of y = A cox + B sinx
Open in App
Solution
Differentiating above equation with respect to x, we get
dydx=−Asinx+Bcosx.
Again differentiating above equation with respect to x, we get
d2ydx2=−Acosx−Bsinx=−(Acosx+Bsinx)=−y.
So, the differential equation is
d2ydx2+y=0