verify that the given differential equation is not exact. (-xy sin(x)+2y cos(x))dx + 2x cos(x)dy = 0. if the…

Explanation:

Step1: Identify M and N

Given the differential equation ((-xy\sin(x)+2y\cos(x))dx + 2x\cos(x)dy = 0), we have (M=-xy\sin(x)+2y\cos(x)) and (N = 2x\cos(x)).

Step2: Calculate (M_y) and (N_x)

Differentiate (M) with respect to (y): [M_y=-x\sin(x) + 2\cos(x)] Differentiate (N) with respect to (x): [N_x=2\cos(x)-2x\sin(x)] Since (M_y\neq N_x), the given differential - equation is not exact.

Step3: Multiply by integrating factor (\mu(x,y)=xy)

The new (M) is (M'=xy(-xy\sin(x)+2y\cos(x))=-x^{2}y^{2}\sin(x)+2xy^{2}\cos(x)) The new (N) is (N'=xy(2x\cos(x)) = 2x^{2}y\cos(x))

Step4: Calculate new (M_y) and (N_x) for the new equation

Differentiate (M') with respect to (y): [M_y'=-2x^{2}y\sin(x)+4xy\cos(x)] Differentiate (N') with respect to (x): [N_x'=4xy\cos(x)-2x^{2}y\sin(x)] Since (M_y' = N_x'), the new equation is exact. To solve the exact equation (M'dx+N'dy = 0), we integrate (M') with respect to (x) and (N') with respect to (y). Integrate (M') with respect to (x): (\int(-x^{2}y^{2}\sin(x)+2xy^{2}\cos(x))dx) Using integration by parts (\int u dv=uv-\int v du), for (\int x^{2}y^{2}\sin(x)dx), let (u = x^{2}y^{2}), (dv=\sin(x)dx), then (du = 2xy^{2}dx), (v=-\cos(x)) (\int x^{2}y^{2}\sin(x)dx=-x^{2}y^{2}\cos(x)+2y^{2}\int x\cos(x)dx) For (\int x\cos(x)dx), let (u = x), (dv=\cos(x)dx), then (du=dx), (v = \sin(x)), so (\int x\cos(x)dx=x\sin(x)+\cos(x)+C) (\int(-x^{2}y^{2}\sin(x)+2xy^{2}\cos(x))dx=x^{2}y^{2}\cos(x)-2y^{2}(x\sin(x)+\cos(x))+y^{2}\int2\cos(x)dx=x^{2}y^{2}\cos(x)-2xy^{2}\sin(x)- 2y^{2}\cos(x)+2xy^{2}\sin(x)=x^{2}y^{2}\cos(x)-2y^{2}\cos(x)+h(y)) Integrate (N') with respect to (y): (\int2x^{2}y\cos(x)dy=x^{2}y^{2}\cos(x)+g(x)) The solution of the exact equation is (x^{2}y^{2}\cos(x)-2y^{2}\cos(x)=C)

Answer:

The solution of the differential equation after using the integrating factor is (x^{2}y^{2}\cos(x)-2y^{2}\cos(x)=C)