Question

In Problems 5-6, determine an Integrating Factor for the given DE. 5. [2x + (x2 + y2) cot x]dx + 2ydy = 0. 6. + (+ 1)dy = 0. 7. Use Euler's Method with the specified Step Size (h), to determine the solution to the given IVP at the specified point. y = x2 + y², y(1) = 2, h = 0.2; y(1.4)~?

Answer 1

Solution:

5.

Given equation is $\small \left [ 2x+\left ( x^{2} +y^{2}\right )cotx \right ]dx+2ydy=0$

Comparing with $\small Mdx+Ndy=0$ , we get

$\small M=2x+\left ( x^{2} +y^{2}\right )cotx$

$\small N=2y$

$\small \therefore \frac{\partial M}{\partial y}=2ycotx,\: \: \: \: \: \: \: \: \: \frac{\partial N}{\partial x}=0$

Since   $\small \frac{\partial M}{\partial y}=2ycotx\neq \frac{\partial N}{\partial x}=0$ , the given equation is not an exact equation.

Now,

$\small \frac{1}{N}\left [ \frac{\partial M}{\partial y}-\frac{\partial N}{\partial x} \right ]=\frac{1}{2y}\left [ 2ycotx-0 \right ]=cotx=f\left ( x \right )$

$\small \therefore e^{\int f\left ( x \right )dx}=e^{\int cotx\: dx}=e^{\log \left ( sinx \right )}=sinx$

$\small \therefore sinx$ is an integrating factor.

6.

Given equation is $\small x^{2}ydx+y\left ( x^{3}+1 \right )dy=0$

Comparing with $\small Mdx+Ndy=0$ , we get

$\small M=x^{2}y$

$\small N=y\left ( x^{3}+1 \right )$

$\small \therefore \frac{\partial M}{\partial y}=x^{2},\: \: \: \: \: \: \: \: \: \frac{\partial N}{\partial x}=3x^{2}y$

Since   $\small \frac{\partial M}{\partial y}=x^{2}\neq \frac{\partial N}{\partial x}=3x^{2}y$ , the given equation is not an exact equation.

Now,

$\small \frac{1}{M}\left [ \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right ]=\frac{1}{x^{2}y}\left [ 3x^{2}y-x^{2} \right ]=3-\frac{1}{y}=f\left ( y \right )$

$\small \therefore e^{\int f\left ( y \right )dy}=e^{\int\left ( 3-\frac{1}{y} \right )\: dy}=e^{3y-\log y}=e^{3y}\cdot e^{\log y^{-1}}=\frac{e^{^{3y}}}{y}$

$\small \therefore \frac{e^{^{3y}}}{y}$ is an integrating factor.

$\small \therefore sinx$ is an integrating factor.