Question

If $\mathbf{xM}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{+}\mathbf{yN}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\equiv \mathbf{0}$, find the solution to the equation $\mathbf{M}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{dx}\mathbf{+}\mathbf{N}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$.

Short answer

$\mathbf{x}\mathbf{=}\mathbf{0}\mathrm{and}\mathbf{y}\mathbf{=}\mathbf{Cx}$

Step-by-step solution

Evaluate the given equation

Given that $\mathbf{xM}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{+}\mathbf{yN}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\equiv \mathbf{0} ......\mathbf{\left(}\mathbf{1}\mathbf{\right)}$,

Which is equivalent to $\mathbf{yN}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{=}\mathbf{-}\mathbf{xM}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right) ......\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Substitute $\mathbf{x}\mathbf{=}\mathbf{0}$to get the function in terms of y.

localid="1664189460946" $\mathbf{yN}\left(\mathbf{0}\mathbf{,}\mathbf{y}\right)\mathbf{=}\mathbf{0}$

So, the result shows that the solution of the given equation is.

Now, take $\mathbf{M}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{dx}\mathbf{+}\mathbf{N}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{dy}\mathbf{=}\mathbf{0} ......\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

Multiply${\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{y}a$ on both sides to get,

${\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{yM}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{dx}\mathbf{+}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{yN}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$

simplification method

Now use the equation (2) here.

$\begin{array}{c}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{yM}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{dx}\mathbf{+}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\left(\mathbf{-}\mathbf{xM}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\right)\mathbf{dy}\mathbf{=}\mathbf{0}\\ {\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{yM}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{dx}\mathbf{-}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{xM}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{dy}\mathbf{=}\mathbf{0}\\ \mathbf{xM}\left({\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{ydx}\mathbf{-}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{dy}\right)\mathbf{=}\mathbf{0}\end{array}$

Observe the founded equation. We get,

$\frac{\mathbf{d}\left({\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{y}\right)}{\mathbf{dx}}\mathbf{=}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{ydx}\mathbf{-}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{dy}$

By using the above equation, we get, $\mathbf{-}\mathbf{xMd}\left({\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{y}\right)\mathbf{=}\mathbf{0}$.

Since we know that neither ofxnorMis zero, then the derivative must be zero.

Hence

$\begin{array}{c}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{y}\mathbf{=}\mathbf{C}\\ \mathbf{y}\mathbf{=}\mathbf{Cx}\end{array}$

Therefore, the solutions for the given equation is $\mathbf{x}\mathbf{=}\mathbf{0}$ and $\mathbf{y}\mathbf{=}\mathbf{Cx}$.