Homework Help: CALCULUS

Posted by Anonymous on Saturday, May 26, 2012 at 2:58pm.

Using separation of variables technique, solve the following differential equation with initial condition dy/dx = (yx + 5x) / (x^2 + 1) and y(3) = 5. The solution is:
a.) y^2 = ln(x^2 + 1) + 25 - ln10
b.) ln(abs(y+5)) = ln(x^2 + 1)
c.) ln(abs(y+5)) = arctan3 + ln10 - arctan3
d.) ln(abs(y+5)) = (1/2)ln(x^2 + 1) + (1/2)ln10
e.) y = ln(x^2 + 1) + 50 - ln10

Based on the initial condition y(3)=5, I know that e cannot be correct. However, I don't know where to start or how to work this out. My main problem is figuring out how to separate the variables.

CALCULUS - Steve, Saturday, May 26, 2012 at 3:48pm
I think D is your answer, but it's a strange way to write it. It just means

y+5 = sqrt(1) * sqrt(x^2+1)

Separation of variables here is rather easy, since you have

dy/dx = x(y+5)/(x^2+1) -- just divide by (y+5) and multiply by dx.

visit wolframalpha . com and type in

solve dy/dx = (yx + 5x) / (x^2 + 1), y(3)=5

it will solve the DE and if you click the Show Steps button, it will show how to separate the variables and all.
CALCULUS - Damon, Saturday, May 26, 2012 at 3:57pm
x(y+5) dx = (x^2+1) dy

x dx/(x^2+1) = dy/(y+5)

(1/2)ln(x^2+1) =
ln(y+5)+ c

if x = 3, y = 5
(1/2) ln 10 = ln 10 + c
c = -(1/2) ln 10

(1/2) ln(x^2 + 1) = ln(y+5) -(1/2) ln 10

(1/2) ln 10(x^2+1) = ln(y+5)

ln 10(x^2+1) = ln(y+5)^2

10(x^2+1) = (y+5)^2

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