Dy/dx=y+3 y(0)=2

Question

y(log2)=?

Answer 1

The First derivative Equation is .

dy = (y +3) dx

Apply integration on both sides.

ln(y+ 3) =x + C

ln(y+3) = x + C

ln(y+3) = x+C

Now y(0) = 2.

ln(2+3) = 0 + C

C = ln(5).

C = 1.609

Substitute C = 1.386 in the Equation ln(y+3) = x+C.

ln(y+3) = x+1.609

y+3 = e^(x + 1.609)

Subtract 2 from both sides.

y= e^(x + 1.609) - 2

Therefore the Equation is modified as .

Now y(log2)

y(log2) = 4.7

Therefore the Equation is and y(log2) = 4.7.