Find the differential equation of y = A cox + B sinx

Open in App

Solution

Differentiating above equation with respect to $x$ , we get $\frac{d y}{d x} = - A s i n x + B c o s x .$ Again differentiating above equation with respect to $x$ , we get $\frac{d^{2} y}{d x^{2}} = - A c o s x - B s i n x = - (A c o s x + B s i n x) = - y .$ So, the differential equation is $\frac{d^{2} y}{d x^{2}} + y = 0$

Suggest Corrections

Similar questions

\frac{d y}{d x} = s i n (x + y) + c o s (x + y)

sec x \frac{d y}{d x} = y + sin x

y = - (sin x + A) + c e^{sin x} .

\frac{d y}{d x} tan y = sin (x + y) + sin (x - y)

\frac{d y}{d x} - y = s i n x

\frac{d y}{d x} + sin \frac{x + y}{2} = sin \frac{x - y}{2}

Join BYJU'S Learning Program