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=R*4.1 Checking the Exactness of a L2-ODE-VC =

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Given
The Given L2-ODE-VC Equation that needs to be tested for its Exactness is

Problem
Check ($$) for the Exactness Conditions of a L2-ODE-VC.

Nomenclature
$$p:=y'(x):=\frac{dy(x)}{dx}$$

Testing for Exactness Conditions of a N2-ODE
Since the given equation does not have a missing y, the equation has to be tested for exactness using the Exactness Conditions for a N2-ODE, which are as below:

1st Exactness Condition:

The ODE should be of the form

2nd Exactness Condition Set:

By observing ($$), we find that it satisfies the 1st Exactness Condition as shown below:

Hence, we have identified that,

Let us now test for the 2nd Exactness Conditions. But before that let us find the partial derivative terms $$ f_{xx}, f_{xy}, f_{xp}, f_y, f_{yy}, f_{yp}, g_{xp}, g_y, g_{yp}$$ and $$g_{pp}$$ required in the conditions individually.

Using ($$), we get,

Using ($$), we get,

Using ($$), we get,

Using ($$), we get,

Using ($$), we get,

Now, plugging ($$), ($$), ($$), ($$), ($$) and ($$) into ($$) which is the 1st equation of the 2nd Exactness Condition Set, we get,

$$(\frac{-1}{4}x^{\frac{-3}{2}})+(2p)(0)+p^2(0)=2+p(0)-3$$

This proves that the given equation is not Exact. However, for a more thorough proof, let us also check the 2nd equation of the 2nd Exactness Condition Set ($$). Plugging ($$), ($$), ($$) and ($$) into ($$), we get,

$$0+p(0)+2(0)=0$$

Although the 2nd equation of the 2nd Exactness Condition Set holds true, since, the first doesn't, the given ODE cannot be deemed exact. Both the 2nd Exactness Conditions have to hold true for the given ODE to be Exact.

Hence, the given L2-ODE-VC is NOT EXACT.

=R*4.2 Determine the integrating factor, first integral and solution of a L2-ODE-VC =

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem
1. Find $$ n,m \in \mathbb{R} $$ such that ($$) is exact.

2. Show that the first integral $$ \phi(x,y,p) $$ is a L1-ODE-VC and is given by ($$).

3. Using the definition $$ p(x) := y'(x) $$ solve ($$) for $$y(x)$$.

Solution
Use the two relations of the 2nd Exactness Condition to solve for the unknown values $$n$$ and $$m$$.

First, multiply the integrating factor $$x^my^n$$ into the ODE ($$.

Now determine each of the required partial derivatives for the first relation of the 2nd Exactness Condition ($$).

Substitute the required partial derivatives into ($$)

$$\left(m-\frac{1}{4}\right)x^{m-\frac{3}{2}}y^n+2p(n)\left(m+\frac{1}{2}\right)x^{m-\frac{1}{2}}y^{n-1}+p^2(n)(n-1)x^{m+\frac{1}{2}}y^{n-2}=...$$

Now get the required partial derivatives for the second relation of the 2nd Exactness Condition($$).

Now substitute the required partial derivatives into ($$):

If $$x$$, $$y$$ $$ \in \mathbb{R} $$ and $$ x \neq 0$$ and $$y \neq 0$$, therefore $$ n=0$$ to satisfy ($$)

Substitute $$n=0$$ into ($$) and solve for $$m$$

The only way for ($$) to hold true is for both coefficients $$\left(m^2-\frac{1}{4}\right)$$ and $$(2m-1)$$ to be equal to zero. This is because the same value raised to different powers cannot be equal i.e. $$x^{m-\frac{3}{2}} \neq x^m$$.


 * $$\left(m^2-\frac{1}{4}\right)=0 \rightarrow m = \pm \frac{1}{2}$$


 * $$ \left(2m-1\right)=0 \rightarrow m = \frac{1}{2}$$


 * $$ m = \frac{1}{2} $$ satisfies both equations so reject $$m=-\frac{1}{2}$$


 * $$ m = \frac{1}{2} \,\,\,\,\,\, n=0 $$

Now that the values for n and m are known, multiply ($$) by the integrating factor $$x^\frac{1}{2}$$ to make it exact.

The function of integration $$h(x,y)$$ can be determined from the definition of $$g(x,y,p)$$

Use the definition $$p(x)=y'(x)$$ to solve ($$) for $$y(x)$$

To divide the coefficient of term $$xy'$$ on both side of ($$), we have,

To compare ($$) with the particular form of ($$), which is now listed as follows,

we have,

To find the integrating factor $$h(x)$$ first using the formula ($$),

Substituting ($$) into ($$) yields,

Then find final solution of $$y(x)$$ using the formula ($$), $$b(x)$$ in ($$) and obtained $$h(x)$$,

Here ($$) is the solution of $$y(x)$$ for ($$)

=R*4.3 Generating a class of exact L2-ODE-VC = On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Given
A L2-ODE-VC :

The first intregal $$ \phi(x,y,p)$$ can also be expressed as:

Problem
Show that($$) and ($$) lead to a general class of exact L2-ODE-VC of the form:

Nomenclature
$$\displaystyle p := y ' := \frac {dy}{dx}$$

Derivation of Eq. 4.3.3
The first exactness condition for L2-ODE-VC:

From ($$) and ($$), we can infer that

Integrating ($$), w.r.t p, we obtain:

Partial derivatives of $$ \phi $$ w.r.t to x and y can be written as:

Substituting the partial derivatives of $$\phi$$ w.r.t x,y and p [($$), ($$), ($$)] into ($$), we obtain:

Comparing ($$) with ($$), we can write:

Thus $$ \displaystyle \frac{\partial k(x,y)}{\partial x} = R(x) y $$

Integrating w.r.t x,

Substituting the $$ k(x,y) $$ obtained in ($$) back into the expression for $$ \phi $$ obtained in ($$), we obtain:

The partial derivative of $$\phi$$ ($$) w.r.t y,

But from ($$) and ($$), we see that $$ \displaystyle \phi_y = Q(x) $$.

So, $$Q(x)=T(x)+k'_1(y)$$

Since, $$ Q(x)$$ is only a function of $$x$$, so, we can now say that $$ T(x) = Q(x) $$ and $$ {k_1}^{\prime}(y) = 0 $$.

Thus $$ k_1(y) $$ is a constant.

Hence we obtain the following expression for $$ \phi $$:

which represents a general class of Exact L2-ODE-VC.

=R*4.4 Solving a L2-ODE-VC = On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem
1. Show that ($$) is exact.

2. Find $$\displaystyle \phi$$

3. Solve for $$\displaystyle y(x)$$

Nomenclature
$$\displaystyle p =: y ' =: \frac {dy}{dx}$$

$$\displaystyle f_{ij} = \frac{d^{2}f}{\partial i \partial j}$$

Exactness Conditions
The exactness conditions for N2-ODE (Non Linear Second Order Differential Equation) are:

First Exactness condition

For an equation to be exact, they must be of the form

$$ \displaystyle G(x,y,y',y' ') = g(x,y,p) + f(x,y,p)y' ' = \frac{\mathrm{d} \phi}{\mathrm{d} x}$$

Second Exactness Condition

Work
We have

$$ \displaystyle G = (cos x)y'' + (x^2 - sin x)y' + 2xy = 0$$

Where we can identify

$$\displaystyle g(x,y,p) = (x^2 - sin x)y' + 2xy$$

and

$$\displaystyle f(x,y,p)= cos x $$

Thus the equation satisfies the first exactness condition.

For the second exactness condition, we first calculate the various partial derivatives of f and g.

$$\displaystyle g_p = x^2 - sinx$$

$$\displaystyle g_{pp} = 0$$

$$\displaystyle g_{xp} = 2x -cosx$$

$$\displaystyle g_{yp} = 0$$

$$\displaystyle g_y = 2x$$

$$\displaystyle f_p = 0$$

$$\displaystyle f_{xp} = f_{yp} = 0$$

$$\displaystyle f_y = f_{yy} = f_{xy} = 0$$

$$\displaystyle f_x = -sinx$$

$$\displaystyle f_{xx} = -cosx $$

Substituting the values in ($$) we get

$$\displaystyle L.H.S = -cosx + 0 + 0 = -cosx$$

$$\displaystyle R.H.S = 2x -cosx + 0 - 2x = -cosx$$

Therefore the first equation satisfies.

Substituting the values in ($$) we get

$$\displaystyle L.H.S. = 0 + 0 + 0 = 0$$

$$\displaystyle R.H.S = 0$$

Therefore the second equation satisfies as well.

Thus the second exactness condition is satisfied and the given differential equation is exact.

Now, we have $$\displaystyle f(x,y,p) = \phi _p $$

Integrating w.r.t. p, we get

$$\displaystyle \phi = \int cosx.dp + h(x,y)$$

where h(x,y) is a function of integration as we integrated only partially w.r.t. p.

Partially differentiating ($$) w.r.t x

$$\displaystyle \phi_x = -psinx. + h_x$$

Partially differentiating ($$) w.r.t y

$$\displaystyle \phi_y = h_y$$

From equation ($$), we have

$$\displaystyle g(x,y,p) = \phi_x + \phi_{y}.y'$$

$$\displaystyle = -p.sinx +h_x + h_{y}.p$$

$$\displaystyle = p(h_y - sinx) + h_x $$

We have established that

$$\displaystyle g(x,y,p) = (x^2 - sin x)y' + 2xy$$

Comparing the two equations, we get,

$$\displaystyle h_x = 2xy$$

On integrating,

$$\displaystyle h = x^{2}y + k_{1}(x)$$

Thus,

$$\displaystyle h_x = 2xy +k_{1}' = 2xy, \therefore k_1 = const $$

Thus we have

$$\displaystyle \phi = cosx.y' + x^{2}y + k_1 = k_2$$

$$\displaystyle \phi = cosx.y' + x^{2}y = k$$

$$\displaystyle y' + \frac{x^{2}y}{cosx} = \frac{k}{cosx}$$

This N1-ODE can be solved using the Integrating Factor Method that we very well know.

$$\displaystyle h(x) = e^{\int \frac{x^{2}}{cosx} .dx} $$

$$\displaystyle y = \frac{1}{h(x)}.\int h(x).\frac {k}{cosx}.dx + c$$

$$\displaystyle y = \frac {1}{e^{\int \frac{x^{2}}{cosx}.dx}} \int \left[e^{\int \frac{x^{2}}{cosx} .dx} {\frac{k}{cosx}}dx \right] + c$$

=R*4.5 Show equivalence to symmetry of second partial derivatives of first integral = On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Given
where

Problem
Show equivalence to symmetry of mixed second partial derivatives of first integral, that is

$$\displaystyle \phi_{xy}=\phi_{yx}, \phi_{py}=\phi_{yp}, \phi_{xp}=\phi_{px} $$

where

$$\displaystyle p(x):=y'(x)$$

Solution
From ($$), we have,

$$\displaystyle \frac{d}{dx}g_1=\frac{d}{dx}[\phi_{xp}+\phi_{yp}y'+\phi_{pp}y'']+\frac{d\phi_y }{dx} $$

Substituting ($$),($$) and ($$)  into ($$)  yields

$$\displaystyle \frac{\partial}{\partial y}(\frac{d \phi}{dx}) -\frac{d}{dx}[\phi_{xp}+\phi_{yp}y'+\phi_{pp}y]-\frac{d }{dx}(\frac{\partial\phi}{\partial y})+\frac{d}{dx}[\phi_{px}+\phi_{py}y'+\phi_{pp}y]=0 $$

Because

$$\displaystyle \frac{\partial}{\partial y}(\frac{d \phi}{dx})=\frac{\partial}{\partial y}(\frac{\partial \phi}{\partial x}+\frac{\partial \phi}{\partial y}\frac{dy}{dx}+\frac{\partial \phi}{\partial y'}\frac{dy'}{dx})=\phi_{xy}+\phi_{yy}\frac{dy}{dx}+\phi_{py}\frac{dy'}{dx} $$

$$\displaystyle \frac{d }{dx}(\frac{\partial\phi}{\partial y}) =[\frac{\partial}{\partial x}(\frac{\partial\phi}{\partial y})+\frac{\partial}{\partial y}(\frac{\partial\phi}{\partial y})\frac{dy}{dx}+\frac{\partial}{\partial y'}(\frac{\partial\phi}{\partial y})\frac{dy'}{dx}] =\phi_{yx}+\phi_{yy}\frac{dy}{dx}+\phi_{yp}\frac{dy'}{dx} $$

Thus

Substituting ($$) into ($$) yields

Because

Substitute ($$) into ($$), we have

$$\displaystyle g_0-\frac{dg_1}{dx}+\frac{d^2g_2}{dx^2}=(\phi_{xy}-\phi_{yx})+2(\phi_{py}-\phi_{yp})y''+\frac{d}{dx}(\phi_{px}-\phi_{xp})+y'\frac{d}{dx}(\phi_{py}-\phi_{yp})=0 $$

Since $$y''$$ and $$y'$$ can be any solution of any second order ODE from which the equation $$g_0-\frac{dg_1}{dx}+\frac{d^2g_2}{dx^2}=0 $$ is established. That is, the factor$$2y''+y'\frac{d}{dx}$$ can be arbitrary. For the left side of ($$) being zero under any circumstances, we should have,

while

From ($$),we obtain,

Define, $$p(x):=y'(x)$$,$$q(x):=y''x)$$.

Substituting the definition above into the ($$) yields

Thus we are now left with ($$) $$\displaystyle (\phi_{xy}-\phi_{yx})+\frac{d}{dx}(\phi_{px}-\phi_{xp})=0 $$

Let us analyze the second term first.

$$\displaystyle \frac{d}{dx}(\phi_{px}-\phi_{xp})$$

This is the ONLY term which will yield terms with $$y' '$$ in it. Thus for the entire equation to be zero, the $$y' '$$ terms must cancel out, and for that, $$\phi_{px}$$ must equal $$\phi_{xp}$$, otherwise we will not get the required equality.

Thus we infer that

Substituting the definition $$p(x):=y'(x)$$ into the ($$) yields

We are now left with $$\displaystyle (\phi_{xy}-\phi_{yx}) = 0$$

Thus

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