+11 For the following separable first order diff eq. I need to solve the initial value problem. y(0)=0
\(\frac{dy}{dx}=\frac{y(x^2+4)}{y^2+1} \)
When separating y and x I get:
\(\int y+\frac{1}{y}dy=\int x^2+4 dx\)
The implicit solution should now be
\(\frac{1}{2}y^2+ln\left |y \right |=\frac{1}{3}x^3+4x+C\)
The problem is now that the initial value problem y(0)=0 can be solved using the above equation since ln(0) is undefined.
Can anyone explain how to solve this problem?
thanks!
+33653 If y(0) = 0 then dy/dx(0) = 0 as well. This means y never changes from zero!
Are you sure y(0) = 0?
.
+11 Yes I'm sure about y(0)=0
Part A was solve for y(0)=1
And in part B I have to solve for y(0)=0
+11 Why does it mean y never changes from zero?
When you fill in y=0 and x=0, in dy/dx don't you only get the change in (0,0) instead of all values for x.
Since y=0 is a solution for y and since I can't fill in 0 in the my implicity solution doesn't this mean my solution is incomplete?
+33653 Here is a numerical solution comparison using initial conditions 0 (yy) and 1 (y)
You can see that although y(0) = 1 gives rise to a changing curve, y(0) doesn't.
+33653 Think about taking a small step, deltax, away from zero. If y(0) = 1 then dy/dx is non zero at x=0 and will start to change the value of y. However, if y(0) = 0, then day/dx is zero and hence is unable to give rise to a change in y, so at deltax the value of y is still zero. Repeating this for another step of deltax therefore still doesn't lead to a change in y. y stays at zero for all x.
