Question

can someone help me solve this differential equation please?

Answer 1

Given the form of the ODE, I would suggest checking your partial derivatives for exactness.
\[\underbrace{\left(y\cos xy+\frac{y}{2x}\right)}_{M(x,y)}\,dx+\underbrace{\left(x\cos xy+\frac{1}{2}\ln x+\frac{1}{e^y}\right)}_{N(x,y)}\,dy=0\]
which has partial derivatives
\[M_y=\cos xy-xy\sin xy+\frac{1}{2x}\\
N_x=\cos xy-xy\sin xy+\frac{1}{2x}\]
Since \(M_y=N_x\), the equation is indeed exact.

Answer 2

I did the antiderivative of both of them and added them together is this correct? : ysin(xy)/x + ylnx/x +xsin(xy)/y +lnx-x/2+e^-y=C

Answer 3

No there's more to the procedure than that. You're looking for a solution of the form \(\Psi(x,y)=C\). Upon differentiating, the chain rule gives you the general form,
\[\underbrace{\frac{\partial}{\partial x}\Psi(x,y)}_{\Psi_x=M(x,y)}+\underbrace{\frac{\partial}{\partial y}\Psi(x,y)}_{\Psi_y=N(x,y)}\frac{dy}{dx}=0\]
What you would do is integrate one of the functions \(M\) or \(N\) and find an expression for \(\Psi\). For instance, proceeding with \(M\),
\[\begin{align*}
\Psi_x&=M(x,y)\\\\
&=y\cos xy+\frac{y}{2x}\\\\
\int \Psi_x\,dx&=\int\left(y\cos xy+\frac{y}{2x}\right)\,dx\\\\
\Psi&=\frac{y\sin xy}{y}+\frac{y}{2}\ln |x|+f(y)\\\\
&=\sin xy+\frac{y}{2}\ln |x|+f(y)
\end{align*}\]
Presumably, you have the condition that \(x\neq0\), or better, that \(x>0\). Now you would differentiate with respect to the other variable, \(y\),
\[\begin{align*}
\Psi_y&=x\cos xy+\frac{1}{2}\ln x+f'(y)\\\\
N(x,y)&=\\\\
x\cos xy+\frac{1}{2}\ln x+\frac{1}{e^y}&=\\\\
e^{-y}&=f'(y)
\end{align*}\]
From here you can solve for \(f(y)\) and plug it into the equation \(\Psi=\sin xy+\frac{y}{2}\ln |x|+f(y)\).