User:Egm6321.f10.team6.lee/HW2

= Problem 6 - Verify condition of Exactness =

Required
Verify that eqn(6.1) satisfy the first condition of exactness(6.2).

where,
 * First condition of exactness

Solution
From the eqn(6.1),

Total differential of function $$\phi\left(x,y\right)$$,

As$$\phi\left(x,y\right)=k=const$$, then $$ d\mathbf{\phi}=0 $$, We can rewrite the eqn(6.5),

For the given problem,

So, we can verify that the eqn(6.1) could be converted to the form of the first condition of exactness(6.2).

= Problem 7 =

Solution to Part I
from (7.17)

We found integrting factor $$ h\left(x\right)$$

Let's make the given eqn(7.3) exact.

Multiply integrting factor $$ h\left(x\right)$$ both side.

then,

Exactness of the new Equation
This equation is exact because it meets two conditions of exactness.
 * First condition of exactness
 * Second condition of exactness

Find $$\phi(x,y)=k$$
As for $$\phi\left(x,y\right)$$

We can find $$\phi\left(x,y\right)=k$$ by integrating $$\bar{M}\left(x,y\right)$$ for x, or $$\bar{N}\left(x,y\right)$$ for y.

i.e.

Therefore,

We can find $$k\left(x\right)$$ by $$\frac{\partial \phi(x,y)}{\partial x}$$

We can find $$ y\left(x\right)$$.

Verify the Exactness
Eqn(7.5) satisfies the first condition of exactness because we can re-order that equation like this.
 * First condition of exactness

This equation is not exact, becuase,
 * Second condition of exactness


 * Integrating Factor

Making the equation exact by multiply Euler integrating factor $$ h\left(x\right) $$.

Exactness of the new Equation
We found integrting factor $$ h\left(x\right)$$

Let's make the given eqn(7.5) exact.

Multiply integrting factor $$ h\left(x\right)$$ both side.

then,

This equation is exact because it meets two conditions of exactness.
 * First condition of exactness
 * Second condition of exactness

Find $$\phi(x,y)=k$$
As for $$\phi\left(x,y\right)$$

We can find $$\phi\left(x,y\right)=k$$ by integrating $$\bar{M}\left(x,y\right)$$ for x, or $$\bar{N}\left(x,y\right)$$ for y.

i.e.

Therefore,

We can find $$k\left(x\right)$$ by $$\frac{\partial \phi(x,y)}{\partial x}$$

We can find $$ y\left(x\right)$$.

= Problem 9 - Exactness =

Required
where,
 * 1. Show the given problem is either exact or can be made exact
 * 2. Find $$\phi\left(x,y\right)=k$$

Verify the Exactness
We can get the differential equation from the given condition.

where,

Eqn(9.4) is satisfied 1st condition of exactness because it has the form of Eqn(9.5)
 * First condition of exactness
 * Second condition of exactness

Before we proceed, we can rewrite the Eqn(9.4) as $$ {e}^{2y}\neq 0 $$

Let's check wheter the equation satisfies the Eqn(9.7)

Eqn(9.6) is not exact.

Let's make the equation exact by multiply Euler integrating factor $$ h\left(x\right) $$.

Multiply Eqn(9.11) to Eqn(9.6)

Actually, we don't need to find $$ h\left(x\right) $$ like this difficult way.

because $$ N\left(x,y\right) $$ and $$ M\left(x,y\right) $$ are the function of x only, by just making the $$ N\left(x\right)=const $$ we can make $$ {M}_{y}\left(x,y\right)={N}_{x}\left(x,y\right)$$

in other words, just multiply $${\color{blue}\frac{1}{\mathrm{sin}x}}$$, we can make the $$ N\left(x,y\right)=1 $$

then, $$ {M}_{y}\left(x,y\right)={N}_{x}\left(x,y\right)=0$$

This is the same integrating factor that we had found at Eqn(9.11)

Let's check the exactness for the Eqn(9.12). Eqn(9.12) is satisfied 1st condition of exactness because it has the form of Eqn(9.5)
 * First condition of exactness


 * Second condition of exactness

As Eqn(9.12) satisfies both 1st and 2nd condition of exacness, it is EXACT ODE.

Find $$\phi(x,y)=k$$
As for $$\phi\left(x,y\right)$$

We can find $$\phi\left(x,y\right)=k$$ by integrating $$M\left(x,y\right)$$ for x, or $$N\left(x,y\right)$$ for y.

i.e.

Therefore,

We can find $$k\left(y\right)$$ by $$\frac{\partial \phi(x,y)}{\partial y}$$