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March 14th, 2011, 12:19 PM   #1
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Differential equations

hey, wondering if this equation I should use substitution in the process or a different method. ie u=x/t

find the general solution of the equation :

(xt)dx/dt = x^2 + t^2

thanks
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March 14th, 2011, 12:40 PM   #2
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Re: Differential equations

Yes, that is an appropriate substitution for this non-linear equation, leading to an implicit solution for x(t).
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March 15th, 2011, 10:41 AM   #3
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Re: Differential equations

Can you check a these please

1. Find the general solutions of the first order equations.
a) y dy/dx=x+2
?y dy =?(x+2) dx
y^2/2 = (1/2 x^2 +2x) +C
y^2 = x^2 +4x + C
y=sqrt (x^2 +4x + C) general solution

b) dy/dx (y^2 -1) / x
(1/y^2) -1 dy/dx = 1/x
?(1/y^2) -1 dy = ?1/x
(1/y^2 -1) dy =?ln(x) +C
y=tan(ln(x)+C) general solution

c)x^2 dy/dx +2xy=2x
d/dx(x^2 y)=2x exact form
?d/dx (x^2 y) dx = ?2x dx
x^2 y = x^2 +C
y=x^2 / x^2 + C/x^2
y=1+Cx^-2 general solution

d) cosx dy/dx -y sinx = 1
d/dx (cosx y) =1 exact form
?d/dx (cosx y) dx =?1 dx
cosx y = x
y= x/cosx + C general solution
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March 15th, 2011, 11:17 AM   #4
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Re: Differential equations

a) You are correct up to the point:



Next I would write it as:



Complete the square on x:







This is the family of "north-south" opening square hyperbolas centered at (-2,0)

When you take the square root of both sides of an equation you need to put a on one side, typically the side opposite the variable you are solving for. As you wrote the solution, you are only getting the top-half of the hyperbolas in the solution space.

b)

Separate variables:



Use partial fraction decomposition on the left side:













c) Correct.

d) You need to put your constant of integration in place before you divide through by cos(x):



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March 15th, 2011, 12:18 PM   #5
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Re: Differential equations

Thanks very much , although I was sure I had done well on those, never mind, your steps are good to follow.
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March 15th, 2011, 12:40 PM   #6
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Re: Differential equations

Ok, so i attempted my next question, hopefully i did better on this one.

Find the particular solution of the first order differential equation

x^2 dy/dx -y = 1 subject to y(2)=2

divide by x^2 to get equation in standard form

dy/dx - 1/x y = 1/x^2

?(x)=e^-?1/x dx =e^-?x^-2 = e^x^-1
multiply through by integrating factor
e^x^-1 dy/dx - x^-2 y e^x^-1 = 1/x^2 e^x^-1

d/dx (e^x^1 y) = 1/x^2 e^x^-1

?d/dx (e^x^-1 y) dx = ?1/x^2 e^x^-1 dx

e^x^-1 y = 1/x^2 e^1/x

e^1/x y = x^-2 e^1/x

y= x^-2 e^1/x +C / e^1/x

y=x^-2 = 1/x^2 + c/e^1/x general solution

y(2)=2
2=1/2^2 + c/e^1/2
2=0.25+c/1.6487
(2-0.25) x 1.6487 =c
1.75 x 1.6487 = c
2.8358 =c

y=1/x^2 + 2.8358 / e^1/x particular solution
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March 15th, 2011, 03:17 PM   #7
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Re: Differential equations

You have:



Putting the equation in standard form, you should get:



Calculate integrating factor:



You have the correct integrating factor, so I assume your equation in standard form just contains a typo, where you have P(x) = 1/x.





When you integrate the right side, you need to see that you initially have so you need to multiply the integral by a negative so that you get:



In our case,









Determine C from initial conditions:





Thus the solution satisfying the given conditions is:

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March 15th, 2011, 05:18 PM   #8
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Re: Differential equations

By the way, here is the working of the problem you gave in the first post:



Divide through by xt:



Use the substitution:

=ut\:\therefore\:\frac {dx}{dt}=u+t\frac{du}{dt}" />

The equation is transformed to:













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March 16th, 2011, 03:26 AM   #9
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Re: Differential equations

Great

what about this one..

x dy/dx -2y = x^4 e^x divide by x^3

x/x^3 dy/dx -2y/x^3 = x^4 e^x/ x^3

x^-2 dy/dx -2x^-3 y = xe^x

d/dx (x^-2 y) =xe^x exact form
integrate wrt x

d/dx(x^-2 y) dx = xe^x dx

x^2 y = xe^x + e^x

y= xe^x + e^x / x^2

y=e^x/x + e^x/x^2 general solution
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March 16th, 2011, 05:48 AM   #10
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Re: Differential equations

When you integrate:



You get:



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