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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(tdy/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 
Global Moderator Joined: Dec 2006 Posts: 19,066 Thanks: 1621  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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