Question

Check my answer (ODE):
xy' + (1/2)y = x^(3/2) with y(1) = 5/2
Find y(4).

My answer: 5

Answer 1

@satellite73

Answer 2

Okay, I would do this one by Integrating Factor technique, so you have \[xy' + \frac{ 1 }{ 2 } y = x^{\frac{ 3 }{ 2 }}\]First Thing would be divide the whole thing by x, so the coefficent of y' is 1. Giving us \[y' + \frac{ 1 }{ 2x }y= x^{\frac{ 1 }{ 2 }}\] Now there is a little trick for this bit, take \[e^{\int\limits_{ }^{ }\frac{ 1 }{ 2x }dx}\]which is \[e^{\ln(x^\frac{ 1 }{ 2 })}\]\[x^\frac{ 1 }{ 2 }\] Now we multiply the whole equation by this, and we get \[x^\frac{ 1 }{ 2 }y'+\frac{ 1 }{ 2x^\frac{ 1 }{ 2 } }= x\]Now we notice that the LHS is actually an expansion (using a product rule derivative of this: \[\frac{ d }{ dx }(x^\frac{ 1 }{ 2 }y)\]So now we can write the equation as \[\frac{ d }{ dx }(x^\frac{ 1 }{ 2 }y)= x\]Now we can integrate bothsides wrt x, to get \[x^\frac{ 1 }{ 2 }y = \frac{ x^2 }{ 2 }+c\]The c being the constant of integration. we are told that y(1)=5/2, so plug this in and we get \[1.\frac{ 5 }{ 2 }= \frac{ 1 }{ 2 }+c\]Meaning our c is 2. So the equation is now\[x^\frac{ 1 }{ 2 }y = \frac{ x^2 }{ 2 }+2\] Put the value of x=4 in, and we get \[4^\frac{ 1 }{ 2 } y=10\] Now I would just make sure to take into account(or show that you have thought of) the possiblity of 4^(1/2) being -2, which would give your answer as plus or minus 5, but to be honest i am not 100% on taking the -ve square root in this occaion, Hope that this helped anyway!

Answer 3

@Nanoman

5 is correct.
\[y(x)=\frac{x^{3/2}}{2}+\frac{2}{\sqrt{x}} \]