Determine functions N(r, y) and M(x, y) so that the following differential equationsare exact(9 (7dr + N(x, y) dy = 0.T2 + y(ii) M(x, y) d + (rey + 2xy +dy = 0

(i) Given differential equation is y12x−12+xx2+ydx+Nx,ydy=0......1 and the differential equation is…

Answered: Determine functions N(r, y) and M (x,…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Determine functions N(r, y) and M(x, y) so that the following differential equations
are exact
(9 (7
dr + N(x, y) dy = 0.
T2 + y
(ii) M(x, y) d + (rey + 2xy +
dy = 0

Expert Solution

Step 1

(i) Given differential equation is $(y^{\frac{1}{2}} x^{- \frac{1}{2}} + \frac{x}{x^{2} + y}) d x + N (x, y) d y = 0 . . . . . . (1)$ and the differential equation is exact.

We have to find the $N (x, y)$ .

Compare the differential equation (1) with $M d x + N d y = 0$ , we get

$M = y^{\frac{1}{2}} x^{- \frac{1}{2}} + \frac{x}{x^{2} + y}$

Since, the given differential equation is exact

Hence, $\frac{\partial N}{\partial x} = \frac{\partial M}{\partial y} . . . . . . (2)$

Now, differentiate $M = y^{\frac{1}{2}} x^{- \frac{1}{2}} + \frac{x}{x^{2} + y}$ partially with respect to $y$

$\begin{matrix} \frac{\partial M}{\partial y} = \frac{\partial}{\partial y} (y^{\frac{1}{2}} x^{- \frac{1}{2}} + \frac{x}{x^{2} + y}) \\ = \frac{\partial}{\partial y} (y^{\frac{1}{2}} x^{- \frac{1}{2}}) + \frac{\partial}{\partial y} (\frac{x}{x^{2} + y}) \\ = \frac{1}{2} y^{- \frac{1}{2}} x^{- \frac{1}{2}} - \frac{x}{{(x^{2} + y)}^{2}} \end{matrix}$

Hence, $\frac{\partial M}{\partial y} = \frac{1}{2} y^{- \frac{1}{2}} x^{- \frac{1}{2}} - \frac{x}{{(x^{2} + y)}^{2}}$

From equation (2), we get

$\frac{\partial N}{\partial x} = \frac{1}{2} y^{- \frac{1}{2}} x^{- \frac{1}{2}} - \frac{x}{{(x^{2} + y)}^{2}}$

Now, integrate $\frac{\partial N}{\partial x} = \frac{1}{2} y^{- \frac{1}{2}} x^{- \frac{1}{2}} - \frac{x}{{(x^{2} + y)}^{2}}$ with respect to $x$

$\begin{matrix} N = \int (\frac{1}{2} y^{- \frac{1}{2}} x^{- \frac{1}{2}} - \frac{x}{{(x^{2} + y)}^{2}}) \\ = \frac{1}{2} y^{- \frac{1}{2}} \int x^{- \frac{1}{2}} d x - \int \frac{x}{{(x^{2} + y)}^{2}} \\ = \frac{1}{2} y^{- \frac{1}{2}} \int x^{- \frac{1}{2}} d x - \int \frac{1}{t^{2}} \frac{d t}{2} (\begin{array}{l} Substitute x^{2} + y = t \\ \Rightarrow x d x = \frac{d t}{2} \end{array}) \\ = \frac{1}{2} y^{- \frac{1}{2}} \cdot \frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} - \frac{1}{2} \cdot \frac{t^{- 2 + 1}}{- 2 + 1} + C \\ = \frac{1}{2} y^{- \frac{1}{2}} \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} - \frac{1}{2} \cdot \frac{t^{- 1}}{- 1} + C \\ = x^{\frac{1}{2}} y^{- \frac{1}{2}} + \frac{1}{2 t} + C \\ = x^{\frac{1}{2}} y^{- \frac{1}{2}} + \frac{1}{2 (x^{2} + y)} + C (Susbtitute t = x^{2} + y) \end{matrix}$