User:EGM6321.f09.Team3.cgs/HW4

= Problem 1 =

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Given
Consider Legendre differential equation 1 Page 14-2 with n=0, such that the homogeneous solution $$ u_{1}(x)=1 $$

Find
Use Reduction of Order method 2 to find $$ u_{2}(x) $$ (the 2nd homegeneous solution)

Solution
We are given the Legendre Equation

$$(1-x^2)y''-2xy'+n(n-1)=0$$

and we are given that

$$u_1(x)=1$$

and

$$ n=0$$

From Class notes P??-? Eq 1. we know that

$$U_{1}=\int{} \frac{c_{1}}{u_{1}(t)}exp(-1\int {} a_1(s)ds)+C$$

Where $$c_1=exp(k)$$ and $$k$$ is an integration constant.

First we need to rewrite Equation 1 in the form $$ y''+a_1y'+a_0y=0$$

$$ y''+ \frac{-2x}{(1-x^2)}y'+(0)y=0 $$

$$a_1=\frac{-2x}{(1-x^2)}$$

$$\int {} a_1(s)ds)=\int {} \frac{-2s}{(1-s^2)}ds =-2\int {} \frac{s}{(1-s^2)} ds$$

$$\int {} a_1(s)ds)=-2*\frac{1}{2}(log(1-t^2)+C=-(log(1-t^2)+C$$

$$U_{1}=\int{} (\frac{c_{1}}{1}exp(-(-log(1-t^2))+C)dt$$

$$U_{1}=\int{} exp(k)(1-t^2)dt$$

$$U_{1}=xe^k-\frac{1}{3}e^kx^3+C$$

From Class notes P??-? Eq 1. we know that

$$u_2=u_1\int\frac{1}{u_1(x)^2}exp(-(\int a_1(s)ds)dt +C$$

$$u_2=\int exp(--log(1-t^2)+C)dt$$

$$u_2=\int ((1-t^2)+C)dt = x-\frac{1}{3}x^3+Cx$$
 * }

= Problem 2 From King et al Page 28 Problem 1.1a =

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Given
$$ u_1=e^x$$ (Equation 1) and $$ (x-1)y''-xy'+y=0$$ (Equation 2)

Show
$$u_1 $$ is a solution of the given function and determine $$u_2 $$

Show all steps and derivations used in determining that method is necessary.

Solution
Step 1 - Determine that Reduction of Order Method 0 is not possible.

There is an $$ a_0y $$ term in Equation 2 so Reduction of Order Method 0 is not possible.

Step 2 - Determine if Equation 2 is exact

Condition 1 - Can Equation 2 be written in the form $$ f(x,y,y')y''+g(x,y,y') $$?

$$ \underbrace{(x-1)}_{f(x,y,y')}y''\underbrace{-xy'+y}_{g(x,y,y')}=0$$

Equation 2 meets Condition 1 for exactness

Condition 2 - Can the following 2 equations be satisfied?

let $$p=y'$$

$$ f_{xx} + 2p f_{xy} + p^2 f_{yy} = g_{xp} + p g_{yp} - g_y $$ (Equation 3)

$$ f_{xp} + p f_{yp} + 2 f_y = g_pp $$ (Equation 4)

$$f(x,y,p)=x-1$$

$$f_x=1 $$

$$f_y=f_p=f_{xx}=f_{xy}=f_{xp}=f_{yy}=f_{yp}=0$$

$$g(x,y,p)=-xp+y $$

$$g_x= -p $$

$$g_{xp}=-1$$

$$g_{xx}=g_{xy}=0$$

$$g_y= 1 $$

$$g_{yy}=g_{yp}=0$$

$$g_p= -x $$

$$g_pp= 0

$$

Substituting these numbers into Equations 3 and 4 produces:

$$ f_{xx} + 2p f_{xy} + p^2 f_{yy} = g_{xp} + p g_{yp} - g_y $$ (Equation 3)

$$ 0 + 2p (0) + p^2 (0) = -1 + p (0) - 1$$

$$ f_{xp} + p f_{yp} + 2 f_y = g_pp $$ (Equation 4)

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

Equation 3 is not satisfied.

Step 3 - Use Integrating Factor Method to determing if there is a function $$h(x,y)$$ such that

$$\phi(x,y,y',y'')=h(x,y) + \int f(x,y,p)dp$$

$$\phi(x,y,y',y'')=h(x,y) + \int (x-1)dp$$

$$\phi(x,y,y',y'')=h(x,y) + xp-p$$

Assume $$ h(x)$$ ($$h_y=0$$)

$$g(x,y,p)=\phi_x+\phi_yp=h_x+p+(0)p=h_x+p$$

$$g(x,y,p)=y-xp$$

For exactness

$$y-xp=h_x+p$$

This is not easy solvable, so we move on to the Trial Solution Method

Step 4 - Use Trial Solution Method to find $$u_1$$

Step 4a - try $$ u_1=e^{rx} $$ (where r is a constant)

$$ u_1=e^{rx} $$

$$ u_1'=re^{rx} $$

$$ u_1''=r^2e^{rx} $$

Substituting u_1 for y in Equation 2 results in

$$(x-1)r^2e^{rx} -xre^{rx}+e^{rx}=0$$ (Equation 5)

Simplified, Equation 5 becomes

$$(x-1)r^2-xr+1=0$$

The only constant root (a root that is not a function of x) that satisfies this equation is $$r=1</math so

u_1=e^x

Now that we have proven that u_1=e^x we use Reduction of Order Method 2 to find u_2

Step 5 Use Reduction of Order Method 2 to find u_2

Assume

y=Uu_1

The general for of the homogeneous L2_ODE_VC is

Substitute values above for y, y' and y'' to obtain:

Re-arrange to group factors of U, U' and U' tp obtain:

Since is homog. the U term is equal to 0 leaving

Equation (6)

Let Z=U'

Equation 6 becomes

Re-arranging obtains

Integrate both sides to obtain

where k is an integration constant

Take the exponential of both sides to obtain

where C=exp(k)

Z=U'

Y=Uu_1

)$$

$$ u_2= u_1\int (\frac{1 }{u_1^2} exp[-\int a_1 (s)ds] )dt$$

Given $$u_1=e^x$$ and $$a_1=\frac{-x}{x-1} $$

$$ u_2= e^x\int (\frac{1 }{e^{2t}} exp[-\int \frac{-x}{x-1} (s)ds] )dt$$

$$ u_2= e^x\int (\frac{1 }{e^{2t}} exp[-(-log (t-1))] )dt$$

$$ u_2= e^x\int (\frac{(t-1) }{e^2t} dt$$

$$ u_2= e^x(\frac{1}{e^{2x}}- \frac{x}{e^{2x}}-\frac{1}{2e^{2x}}) $$

$$ u_2= \frac{1}{2e^{x}}-x $$


 * }

Given
$$ u_1=x^{-1}sin x$$ and $$ xy''+2y'+xy=0$$

Find
m & n such that the ODE is exact.

Solution
Step 1 - Determine that Reduction of Order Method 0 is not possible.

There is an $$ a_0y $$ term in Equation 2 so Reduction of Order Method 0 is not possible.

Step 2 - Determine if Equation 2 is exact

Condition 1 - Can Equation 2 be written in the form $$ f(x,y,y')y''+g(x,y,y') $$?

$$ \underbrace{x}_{f(x,y,y')}y''+\underbrace{2y'+xy}_{g(x,y,y')}=0$$

Equation 2 meets Condition 1 for exactness

Condition 2 - Can the following 2 equations be satisfied?

let $$p=y'$$

$$ f_{xx} + 2p f_{xy} + p^2 f_{yy} = g_{xp} + p g_{yp} - g_y $$ (Equation 3)

$$ f_{xp} + p f_{yp} + 2 f_y = g_pp $$ (Equation 4)

$$f(x,y,p)=x$$

$$f_x=1 $$

$$f_y=f_p=f_{xx}=f_{xy}=f_{xp}=f_{yy}=f_{yp}=0$$

$$g(x,y,p)=2p+xy $$

$$g_x= y $$

$$g_{xy}=1$$

$$g_{xx}=g_{xp}=0$$

$$g_y= x $$

$$g_{yy}=g_{yp}=0$$

$$g_p= 2 $$

$$g_pp= 0

$$

Substituting these numbers into Equations 3 and 4 produces:

$$ f_{xx} + 2p f_{xy} + p^2 f_{yy} = g_{xp} + p g_{yp} - g_y $$ (Equation 3)

$$ 0 + 2p (0) + p^2 (0) = -1 + p (0) - x$$

$$ f_{xp} + p f_{yp} + 2 f_y = g_pp $$ (Equation 4)

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

Equation 3 is not satisfied.

Step 3 - Use Integrating Factor Method to determing if there is a function $$h(x,y)$$ such that

$$\phi(x,y,y',y'')=h(x,y) + \int f(x,y,p)dp$$

$$\phi(x,y,y',y'')=h(x,y) + \int (x)dp$$

$$\phi(x,y,y',y'')=h(x,y) + xp$$

Assume $$ h(x)$$ ($$h_y=0$$)

$$g(x,y,p)=\phi_x+\phi_yp=h_x+p+(0)p=h_x+p$$

$$g(x,y,p)=y-xp$$

For exactness

$$y-xp=h_x+p$$

This is not easy solvable, so we move on to the Trial Solution Method

Step 4 - Use Trial Solution Method to find $$u_1$$

Step 4a - try $$ u_1=e^{rx} $$ (where r is a constant)

$$ u_1=e^{rx} $$

$$ u_1'=re^{rx} $$

$$ u_1''=r^2e^{rx} $$

Substituting u_1 for y in Equation 2 results in

$$xr^2e^{rx} +2re^{rx}+xe^{rx}=0$$ (Equation 5)

Simplified, Equation 5 becomes

$$xr^2 +2r+x=0$$

There is no constant root (a root that is not a function of x) that satisfies this equation so

Step 4b - try $$ u_1=xe^{rx} $$ (where r is a constant)

$$ u_1=xe^{rx} $$

$$ u_1'=rxe^{rx}+e^{rx} $$

$$ u_1''=r^2xe^{rx}+ re^{rx}+re^{rx}= r^2xe^{rx}+ 2re^{rx} $$

Substituting u_1 for y in Equation 2 results in

$$r^2x^2e^{rx}+ 2rxe^{rx}+2rxe^{rx}+2e^{rx}+x^2e^{rx}=r^2x^2e^{rx}+ 4rxe^{rx}+2e^{rx}+x^2e^{rx}=0$$ (Equation 6)

Simplified, Equation 6 becomes

$$r^2x^2+ 4rx+2+x^2=x^2(r^2+1)+4rx+2=0$$

There is no constant root (a root that is not a function of x) that satisfies this equation so

Step 4c - try $$ u_1=\frac{e^{rx}}{x} $$ (where r is a constant)

$$ u_1=\frac{e^{rx}}{x} $$

$$ u_1'=\frac {rxe^{rx}-e^{rx}}{x^2}=\frac{re^{rx}}{x}-\frac{e^{rx}}{x^2} $$

$$ u_1''=\frac {r^2xe^{rx}-re^{rx}}{x^2}-\frac{x^2re^{rx}}{x^4}+\frac{2xe^{rx}}{x^4}$$

$$ u_1''=\frac {r^2e^{rx}}{x}-\frac {re^{rx}}{x^2}-\frac{re^{rx}}{x^2}+\frac{2e^{rx}}{x^3}=\frac {r^2e^{rx}}{x}-\frac {2re^{rx}}{x^2}+\frac{2e^{rx}}{x^3}$$

Substituting u_1 for y in Equation 2 results in

$$x ( \frac {r^2e^{rx}}{x}-\frac {2re^{rx}}{x^2}+\frac{2e^{rx}}{x^3}) +2(\frac{re^{rx}}{x}-\frac{e^{rx}}{x^2})+x \frac{e^{rx}}{x} =0$$

$$\frac {r^2e^{rx}}-\frac {2re^{rx}}{x}+\frac{2e^{rx}}{x^2}+\frac{2re^{rx}}{x}-\frac{2e^{rx}}{x^2}+e^{rx} =0$$

$$\frac {r^2e^{rx}} +e^{rx} =0$$(Equation 7) Simplified, Equation 7 becomes

$$r^2 +1=0$$

The root of this equation is $$\sqrt{-1}$$ Which makes $$u_1=e^{ix}$$

From DeMoivre's theorem we know that $$e^{ix}= cos( x) +isin(x)$$

\big_{I DON'T KNOW HOW TO GET TO SIN X/X FROM HERE}

Now that we have proven that $$u_1=\frac{sin x}{x}$$ we use Reduction of Order Method 2 to find $$u_2$$

Step 5 Use Reduction of Order Method 2 to find $$u_2$$

Assume

$$y=Uu_1$$

$$y'=U'u_1 + Uu_1'$$

$$y=Uu_1 + U'u_1'+U'u_1' + Uu_1''$$

$$y=Uu_1 + 2U'u_1'+Uu_1''$$

The general for of the homogeneous L2_ODE_VC is

$$y''+a_1y'+a_0y=0$$

Substitute values above for y, y' and y'' to obtain:

$$(Uu_1 + 2U'u_1'+Uu_1)+a_1(U'u_1 + Uu_1')+a_0Uu_1=0$$

Re-arrange to group factors of U, U' and U' tp obtain:

$$U(u_1+a_1u_1'+a_0u_1)+U'(2u_1'+a_1 u_1)+Uu_1=0$$

Since $$u_1''+a_1u_1'+a_0u_1$$ is homog. the U term is equal to 0 leaving

$$U'(2u_1'+a_1 u_1)+U''u_1=0$$ Equation (6)

Let $$Z=U'$$

Equation 6 becomes

$$Z(2u_1'+a_1 u_1)+Z'u_1=0$$

Re-arranging obtains

$$\frac{Z'}{Z}=-\frac{1}{u_1}(2u_1'+a_1 u_1)$$

$$\frac{Z'}{Z}=-\frac{2u_1'}{u_1}-\frac{a_1 u_1}{u_1}$$

$$\frac{Z'}{Z}=-\frac{2u_1'}{u_1}-a_1$$

Integrate both sides to obtain

$$log (Z)= -2 log (u_1) -\int a_1 (s)ds +k $$ where k is an integration constant

Take the exponential of both sides to obtain

$$exp(log (Z))= exp(-2 log (u_1) -\int a_1 (s)ds +k )$$

$$Z= \frac{C }{u_1^2} exp[-\int a_1 (s)ds] $$ where $$C=exp(k)$$

<Z=U'

$$U=\int (\frac{C }{u_1^2} exp[-\int a_1 (s)ds] ) +C_1$$

$$Y=Uu_1$$

$$Y=u_1(\int (\frac{C }{u_1^2} exp[-\int a_1 (s)ds] ) +C_1)$$

$$Y=C\underbrace{u_1\int (\frac{1 }{u_1^2} exp[-\int a_1 (s)ds] )}_{u_2} +C_1u_1$$)

$$ u_2= u_1\int (\frac{1 }{u_1^2} exp[-\int a_1 (s)ds] )dt$$

Given $$u_1=\frac{sin x}{x} $$ and $$a_1=\frac{2}{x} $$

$$ u_2= \frac{sin x}{x}\int (\frac{t^2 }{ sin^2 t} exp[-\int \frac{2}{s} (s)ds] )dt$$

$$ u_2= \frac{sin x}{x}\int (\frac{t^2 }{ sin^2 t } exp[-(2log (t))] )dt$$

$$ u_2= \frac{sin x}{x}\int (\frac{t^2 }{ sin^2 t }\frac{1}{t^2} dt$$

$$ u_2= \frac{sin x}{x}\int (\frac{1 }{sin^2 t } dt$$

$$ u_2= \frac{sin x}{x}(-cot x)=\frac{sin x}{x}\frac{-cos x}{sin x}=$$

$$ u_2=\frac{-cos x}{x}=$$


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= Problem 4: King 1.3.a =

Given
$$y'' - 2y' + y = x^{3/2}e^x$$

Find
The general solution of $$y(x) = y_H(x) + y_P(x)$$

Solution
First, solve homogeneous equation $$y'' - 2y' + y = 0$$

Reduction of Order Method 0 is not possible, move to step 2

Exactness Condition 1 - Can Equation 2 be written in the form $$ f(x,y,y')y''+g(x,y,y') $$?

Lte p=y'

$$ f(x,y,p)=1 $$

$$ g(x,y,p)=-2p+y $$?

Condition 1 is satisfied, check for Exactness Condition 2

Exactness Condition 2 - Can the following 2 equations be satisfied?

$$

f_{xx} + 2p f_{xy} + p^2 f_{yy} = g_{xp} + p g_{yp} - g_y $$ (Equation 3)

$$ f_{xp} + p f_{yp} + 2 f_y = g_pp $$ (Equation 4)

$$f_x=f_xx=f_xy=f_xp=f_y=f_yy=f_yp=0$$

$$g_y=1$$

$$g_p=-2$$

$$g_x=g_xx=g_xy=g_xp=g_yy=g_yp=g_pp=0$$

If both equations are satisfied, solve for y

$$

0 + 2p (0) + p^2(0) = 0 + p (0) - 1$$

Equation is not satisfied, use Trial Solution Method to find $$u_1$$

Try $$ u_1=e^{rx} $$ (where r is a constant)

Substitute $$u_1$$,$$u_1'$$, and $$u_1''$$ for y in Equation 2

$$ u_1=e^{rx} $$

$$ u_1'=re^{rx} $$

$$ u_1''=r^2e^{rx} $$

Substituting u_1 for y in Equation 2 results in

$$r^2e^{rx} - 2re^{rx} + e^{rx} = 0$$

Simplified, Equation 5 becomes

$$r^2 - 2r + 1 = 0$$

There are 2 roots at $$r=1$$ so $$ u_1=e^{x} $$

Recall that $$a_1=-2$$

$$ u_2= u_1\int (\frac{1 }{u_1^2} exp[-\int a_1 (s)ds] )dt$$

$$ u_2= e^{x}\int (\frac{1 }{e^{2t}} exp[-\int -2 (ds] )dt$$

$$ u_2= e^{x}\int (\frac{1 }{e^{2t}} e^{2t} dt$$

$$ u_2= xe^{x}

Solve equation for u_2then solve for y using the following equation:

To solve for particular solution y_P use the following relationship \bar{W}=

det so are linearly independent.

= Problem 5 - Method of Solving L2-ODE-VC=

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on "show" to see the problem solution

Given
Some function (Equation 1)

Solve
Solve for y(x)

Solution
Step 1 - Determine whether Reduction of Order Method 0 is possible.

If Reduction of Order Method 0 is not possible, move to step 2

Step 2 - Determine if Equation 2 is exact

Step 2a - Exactness Condition 1

Condition 1 - Can Equation 2 be written in the form ?

If Condition 1 is satisfied, check for Exactness Condition 2 (Step 2b). If not, move on to Step 3.

Step 2b - Exactness Condition 2

Condition 2 - Can the following 2 equations be satisfied?

let p=y'

(Equation 3)

(Equation 4)

If both equations are satisfied, solve for y

If either equation is not satisfied move to Step 3

Step 3 - Use Integrating Factor Method to determing if there is a function h(x,y) such that

Assume h(x) (h_y=0)

For exactness

If it is solveable, check for exactess of the new equation. If the new equation is exact, solve for y. If not, move on to the Trial Solution Method

Step 4 - Use Trial Solution Method to find u_1

Step 4a - try u_1=e^{rx} (where r is a constant)

Substitute u_1,u_1', and u_1'' for y in Equation 2

Simplify equation and determine if there is a root r such that r is a constant (not a function of x)

If there is no constant root (a root that is not a function of x) that satisfies this equation go to Step 4b

Step 4b - try u_1=xe^{rx} (where r is a constant)

Substitute u_1,u_1', and u_1'' for y in Equation 2

Simplify equation and determine if there is a root r such that r is a constant (not a function of x)

If there is no constant root (a root that is not a function of x) that satisfies this equation go to Step 4c

Step 4c - try (where r is a constant)

Substituting u_1,u_1', and u_1'' for y in Equation 2 results in

Substitute u_1,u_1', and u_1'' for y in Equation 2

Simplify equation and determine if there is a root r such that r is a constant (not a function of x)

If there is no constant root (a root that is not a function of x) that satisfies this equation go to Step 4d

Step 4d - try another  Keep trying until a function with a constant root is found.

Now that u_1(x) has been found, use Reduction of Order Method 2 to find u_2 (Step 5)

Step 5 Use Reduction of Order Method 2 to find u_2

Assume

y=Uu_1

The general for of the homogeneous L2_ODE_VC is

Substitute values above for y, y' and y'' to obtain:

Re-arrange to group factors of U, U' and U' tp obtain:

Since is homog. the U term is equal to 0 leaving

Equation (6)

Let Z=U'

Equation 6 becomes

Re-arranging obtains

Integrate both sides to obtain

where k is an integration constant

Take the exponential of both sides to obtain

where C=exp(k)

<Z=U'$$

$$U=\int (\frac{C }{u_1^2} exp[-\int a_1 (s)ds] ) +C_1$$

$$Y=Uu_1$$

$$Y=u_1(\int (\frac{C }{u_1^2} exp[-\int a_1 (s)ds] ) +C_1)$$

$$Y=C\underbrace{u_1\int (\frac{1 }{u_1^2} exp[-\int a_1 (s)ds] )}_{u_2} +C_1u_1$$)

$$ u_2= u_1\int (\frac{1 }{u_1^2} exp[-\int a_1 (s)ds] )dt$$

Solve equation for $$u_2 $$then solve for y using the following equation:

$$y=c_1u_1+c_2u_2$$


 * }

Given
$$ u_1=x^{-1}sin x$$ and $$ xy''+2y'+xy=x$$

Find
Solution for y(x)

Solution
From Problem 2 we have already determined that

$$y_H=c_1u_1+c_2u_2=c_1 \frac{sin x}{x} -c_2 \frac{cos x}{x} $$

To solve for particular solution $$y_P$$ use the following relationship $$

\bar{W}= $$ $$ \begin{bmatrix} u_1 & u_2\\ u_1' & u_2'\end{bmatrix}= $$

$$u_1=\frac{sin x}{x} $$

$$u_1'=\frac{xcos x-sin x}{x^2} $$

$$u_2= -\frac{cos x}{x} $$

$$u_2=-\frac{x(-sin x)-cos x}{x^2}= \frac{x(sin x)+cos x}{x^2} $$

$$ \begin{bmatrix} \frac{sin x}{x} & -\frac{cos x}{x}\\ \frac{xcos x-sin x}{x^2} & \frac{x(sin x)+cos x}{x^2}\end{bmatrix} $$

$$W=\text{det}\bar{W}= \frac{sin x}{x}(\frac{x(sin x)+cos x}{x^2})-\frac{xcos x-sin x}{x^2}(-\frac{cos x}{x})$$

$$W= \frac{xsin^2 x+cos x sinx +xcos^2 x-sin x cos x} {x^3} =\frac{x(sin^2x+cos^2x)}{x^3}=\frac{1}{x^2}$$

det$$\bar{W}\neq0$$ so $$u_1 \text{and} u_2$$ are linearly independent.

$$ \begin{bmatrix} c_1'\\ c_2'\end{bmatrix}=\frac{1}{W} \begin{bmatrix} -u_2f\\ u_1f\end{bmatrix}$$

$$c_1'=\frac{1}{W}(-u_2)(f)={x^2}(-\frac{-cos x}{x})(x)=x^2 cos x $$

$$ c_2'=\frac{1}{W}(-u_2)(f)={x^2}(\frac{sin x}{x})(x)=x^2 sin x $$

$$ c_1= \int x^2 cos x =2x cos x+(x^2-2) sin x $$

$$ c_2= \int x^2 sin x=2x sin x-(x^2-2) cos x $$

$$y(x)=y_H+y_P=c_1e^{x} +c_2xe^{x}+(2x cos x+(x^2-2) sin x)e^{x} +(2x sin x-(x^2-2) cos x)xe^{x} $$

$$y(x)=y_H+y_P=c_1e^{x} +c_2xe^{x}+e^xsinx(x^3-3x^2-2) $$

= Problem 7 =

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on "show" to see the problem solution

Given
$$ Homework goes here $$

Find
m & n such that the ODE is exact.

Solution
= Problem 8 =
 * }

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on "show" to see the problem solution

Given
$$ Homework goes here $$

Find
m & n such that the ODE is exact.

Solution
= Problem 9 =
 * }

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on "show" to see the problem solution

Given
$$ Homework goes here $$

Find
m & n such that the ODE is exact.

Solution
= Problem 10 =
 * }

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on "show" to see the problem solution

Given
$$ Homework goes here $$

Find
m & n such that the ODE is exact.

Solution
= Problem 11 =
 * }

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on "show" to see the problem solution

Given
$$ Homework goes here $$

Find
m & n such that the ODE is exact.

Solution
= Problem 12 =
 * }

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on "show" to see the problem solution

Given
$$ Homework goes here $$

Find
m & n such that the ODE is exact.

Solution
= Contributing Members =
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