The solution to the initial value problemdydx=(x−2)(y−3)2, y(0)=0,will always be less than 3; that is, y(x)&lt;3 for x≥0.

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The solution to the initial value problemdydx=(x−2)(y−3)2, y(0)=0,will always be less than 3; that is, y(x)&amp;lt;3 for x≥0.

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Step 1Assume that there exists x0&amp;gt;0 such that y(x0)&amp;gt;3. Then, as y is continuous and y(0)=0, we get by the intermediate value theorem that there exists x1&amp;gt;0 such that y(x1)=3. Now we see that y solves the IVP{dydx=(x−2)(y−3)2;y(x1)=3.However, also u≡3 solves the IVP above. Hence, by local uniqueness we know that there exists ε&amp;gt;0 such that y(x)=3 for x∈(x1−ε,x1+ε)Now we want to show that in fact y(x)=3 for all x≥0. For this we defineS={x∈R : 0≤x≤x1,y(x)=3}.We want to show that S=[0; x1]. For this it is enough to show that S is both open, closed and nonempty. Nonempty is clear as x1∈S. Closed follows from continuity of y and openness follows from redoing the argument above with the local uniqueness theorem.Hence, we have shown that once y attains 3, it has always been equal to 3 (and in fact it would also always be 3 in the future). Thus, this yields a contradiction to the initial condition y(0)=0.Note that we have not used the particular form of the ODE. If you are given any ODE dydx=F(x,y) with F locally lipschitz with F(⋅,y0)=0, then the very same argument as above tell us that if y attains y0, then the solution must be constant for all times.

The solution to the initial value problem \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\left({x}-{2}\right)}{\left({y}-{3}\right)}^{{{2}}},\

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