Initial value problem: y' = 2x(5y - 4) ; y(0) = 1 .?

Question

Initial value problem: y' = 2x(5y - 4) ; y(0) = 1 .?

e^f(x) can be written equivalently as exp f(x) or as e^[f(x)].

answers....

a. y = (1/2)[3 exp(2x^2) - 1]

b. y = (1/3)[5 exp(3x^2) - 1]

c. y = (1/4)[7 exp(4x^2) - 1]

d. y = (1/5)[9 exp(5x^2) - 1]

e. y = (1/2)[exp(2x^2) + 1]

f. y = (1/3)[exp(3x^2) + 2]

g. y = (1/4)[exp(4x^2) + 3]

h. y = (1/5)[exp(5x^2) + 4]

i. y = (1/6)[exp(6x^2) + 5]

j. none of these

Answer 1

Rewrite the expression as:
dy/dx = 2x(5y-4)

Get all the y terms on the left, x terms on the right
dy/(5y-4) = 2xdx

Integrate both sides
(1/5)ln(5y-4) = x^2 + C
ln(5y-4) = 5x^2 + 5C
Solve for y

5y-4 = e^(5x^2 + 5C)
5y = e^(5x^2 + 5C) + 4
y = (e^(5x^2 + 5C) + 4)/5

We know that y(0) = 1, so we can solve for C
1 = (e^(5*0^2 + 5C) + 4)/5
1 = (e^(5C) + 4)/5
5 = e^(5C) + 4
1 = e^(5C)
C = 0

y = (e^(5x^2) + 4)/5

So the answer should be (h)

Answer 2

dy/dx=2x(5y-4)

dy/(5y-4)=2xdx

integrate both sides

1/5ln(|5y-4|)=x^2+c

ln((5y-4)^1/5)=x^2+c

(5y-4)^1/5=Ce^(x^2)
5y-4=De^(5x^2)+4
y=1/5(De^(5x^2)+4)
1=1/5((De^0)+4)
1=1/5(D+4)

5=D+4
D=1

so Y=1/5(e^5x^2+4) (H)