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March 23rd, 2016, 03:57 AM   #1
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Particular solution differential equations

Hi there in major help with the questions attached to this.
Need to determine the particular solution of the following differential equations:
3t(t-dy/dt)=7. Given that y=4 when t=2


1/e^x + 5= x+5 dy/dx. Given that y=2 when x=0

So stuck with these and don't know what to do.image.jpg
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March 23rd, 2016, 04:53 AM   #2
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The differential equation in the second problem can be integrated immediately by integrating each term. Can you give that a try and post your work?
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March 26th, 2016, 04:17 AM   #3
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Looks pretty much like basic Calculus. I would start by writing the equation as 3dy/dt= 3t^2- 7 so that, in differential form, 3dy= (3t^2- 7)dt and integrating both sides 3y= t^3- 6t+ C. Since y(2)= 4, 3(4)= 2^3- 6(2)+ C so 12= 8- 12+ C and C= 16. y(t)= (t^3- 6y+16)/3.
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April 26th, 2016, 06:30 AM   #4
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Originally Posted by harrypalmer101 View Post
3t(t-dy/dt)=7. Given that y=4 when t=2
I'll assume that t > 0, then dividing by t gives 3t - 3dy/dt = 7/t.
Integrating gives 3t²/2 - 3y = 7ln(t) + C, where C is a constant.
Substituting t = 2 and y = 4 gives 6 - 12 = 7ln(2) + C, so C = -6 - 7ln(2).
Hence y = t²/2 - (7/3)ln(t/2) + 2.
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