- #1

- 10

- 0

(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?

- Thread starter deryk
- Start date

- #1

- 10

- 0

(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?

- #2

HallsofIvy

Science Advisor

Homework Helper

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- #3

- 25

- 0

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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