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(2+x^2)dy/dx - x(y+3)= 0 , y(0)=-1

  • Thread starter deryk
  • Start date
10
0
The question asked to solve this initial value problem

(2+x^2)dy/dx = x(y+3)

(2+x^2)dy/(dx*(y+3) = x

dy/[(y+3)*dx]= x/(2+x^2)

dy/(y+3) = x*dx/(2+x^2)

S dy/(y+3) = S x/(2+x^2)*dx

ln(y+3) = ?

Im not sure what to do with the RHS do I do integration by parts?
 

HallsofIvy

Science Advisor
41,625
822
Did it occur to you that, since you have xdx in the numerator, you might try a substitution: u= 2+ x2?
 
well your right so far

so your only problem is the rhs?
if so
let u=x^2+2
du/dx=2x
du=2xdx
dx=du/(2x)

the sub back into rhs as u=x^2+2 and dx=du/(2x)to be left with

x/(u)*du/(2x)
x's cancel
1/u*du
then intergrate with respect to u, once done that sub u=x^2+2
then i'm sure you can find the constant by using y(0)=-1

only way to know how to do this is to practise many examples!
 

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