Question

Given the initial condition y(1)=4, find the particular solution of the equation y'=y(2x+3)

Answer 1

I've solved it up to here:

Dy/dx=y(2x+3)
F(x)=2x+3
G(y)=y
Integral(F(x))=Integral(1/G(y))
X^2+3x=Ln(|y|)+C_1

Answer 2

backward lny = x^2+3x +C

Answer 3

Oh, the C is on the left side?

Answer 4

Ok, so

Answer 5

Ln(|y|)+C=X^2+3x

Answer 6

Now we solve for C?

Answer 7

Oh, I think that's been my mistake. I haven't been solving C last.

Answer 8

y=e^(x^2+3x-C)
2=e^(1^2+3*1-C)

Answer 9

you can solve for C there, since you don't have particular solution. :)

Answer 10

oh, found another mistake. y should equal 4 in the intial solution

Answer 11

It seems that when I solve for C, I get 4-2*ln(2).
Equation I solved: 4=e^(1^2+3*1-C)

Answer 12

\(lny = x^2+3x+C\\y = e^{x^2+3x+C}= C e^{x^2+3x}\\C = \dfrac{y}{e^{x^2+3x}}\)

Answer 13

I have the initial condition though

Answer 14

now, plug x =1,y =4 in to get C

Answer 15

~0.699?

Answer 16

I don't know!! to me, I let it as it is 4e^(-4)

Answer 17

My choices for the particular solution are
y=e^(x^2+3x) - 0.073
y=0.073e^(x^2+3x)
y=e^(x^2+3x+0.073)
y=e^(x^2+3x)+0.073

Answer 18

Oh, I must've messed up when solving for C. How did you get 4/e^4?