91Ó°ÊÓ

The solution to the IVP \[\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ = (x - 2)(y - 3}}{{\bf{)}}^{\bf{2}}}{\bf{,}}\;{\bf{y(0) = 0}}\] will always less than 3.That is \[{\bf{y(x) < 3}}\]for \[{\bf{x}} \ge 0\].

Here the term \[{{\bf{(y - 3)}}^{\bf{2}}}\] it does not contribute to the sign of the derivatives, it is always positive, hence it only contributes to the magnitude. And this is equal to y = 3. So that the function will remain at 3. There is no change the derivatives will remain at zero and nothing will get it out of there.

• Here the term (x- 2), tell the sign of derivatives. Since y(0) = 0, the derivative has a negative sign.
• It will keep a negative sign till the x value reaches 2 after the slope has a positive value.
• So, the function increases. However, the function starts from a value less than 3so, it increases it will have to pass by y = 3, a value which can’t cross and will remain trapped either 3 or at y = 3 itself.

Therefore this statement is true.