(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I'm working on trying to first to solve an ODE on the form

[tex]\sqrt{x(t)}\frac{dx}{dt} = 2\cdot t^2[/tex]

My task is to find a solution on the form [tex]x(t) =nt^s[/tex]

3. The attempt at a solution

I do some seperation and I get that the above is equal to

[tex]\sqrt{x(t)} dx = 2 \cdot t^2 dt[/tex]

Which by integrating on both sides of the equality yields

[tex]\frac{2}{3} x(t)^{3/2} = \frac{2}{3}t^3[/tex]

I mulitiply by a factor of 3 on both sides and divide by a factor of 2 and arrive at

and get [tex]x(t)^{3/2} = t^3[/tex]

which gives the solution

[tex]ln(x(t)) = 2 \cdot ln(t) \Rightarrow x(t) = t^2[/tex]

Then the bad not so easy second question. What is the maximum interval on which x(t) is defined?

According to my textbook I have to equal the two antiderivatives of the original equation to obtain the first endpoint.

and then take the composite of the invers of the anti-derivative of lhs of the original ODE and the antiderivative of the rhs to find the second endpoint.

I then obtain an interval [tex]I = [t^2, sign(t)][/tex]

Have I understood this correctly?

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# Homework Help: Can I express the maximal solution interval of an ODE as

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