Question

verify that the given differential equation is not exact. (-xy sin(x)+2y cos(x))dx + 2x cos(x)dy = 0. if the given de is written in the form m(x,y)dx + n(x,y)dy = 0, one has m_y = n_x = since m_y and n_x --select-- equal, the equation is not exact. multiply the given differential equation by the integrating factor μ(x,y)=xy and verify that the new equation is exact. if the new de is written in the form m(x,y)dx + n(x,y)dy = 0, one has m_y = n_x = since m_y and n_x --select-- equal, the equation is exact. solve.

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 \(M_y
eq N_x\), 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 \(x^{2}y^{2}\cos(x)-2y^{2}\cos(x)=C\)