Chapter 2: Q18E (page 70)

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

Short Answer

Expert verified

$x = 0 and y = Cx$

Step by step solution

Evaluate the given equation

Given that $xM (x, y) + yN (x, y) \equiv 0 . . . . . . (1)$ ,

Which is equivalent to $yN (x, y) = - xM (x, y) . . . . . . (2)$

Substitute $x = 0$ to get the function in terms of y.

localid="1664189460946" $yN (0, y) = 0$

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

Now, take $M (x, y) dx + N (x, y) dy = 0 . . . . . . (3)$

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

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

simplification method

Now use the equation (2) here.

$\begin{matrix} x^{- 1} yM (x, y) dx + x^{- 1} (- xM (x, y)) dy = 0 \\ x^{- 1} yM (x, y) dx - x^{- 1} xM (x, y) dy = 0 \\ xM (x^{- 1} ydx - x^{- 1} dy) = 0 \end{matrix}$

Observe the founded equation. We get,

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

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

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

Hence

$\begin{matrix} x^{- 1} y = C \\ y = Cx \end{matrix}$

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

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

Short Answer

Step by step solution

Evaluate the given equation

simplification method

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If xMx,y+yNx,y≡0, find the solution to the equation Mx,ydx+Nx,ydy=0.

Short Answer

Step by step solution

Evaluate the given equation

simplification method

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

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