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 March 23rd, 2016, 03:57 AM #1 Newbie   Joined: Mar 2016 From: london Posts: 1 Thanks: 0 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 And 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
 March 23rd, 2016, 04:53 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,066 Thanks: 1621 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?
 March 26th, 2016, 04:17 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,199 Thanks: 872 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.
April 26th, 2016, 06:30 AM   #4
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Quote:
 Originally Posted by harrypalmer101 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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