solve the following differential equation dy/dx=x[(x^2 + 4)^(1/2)] with initial condition y(0)=1

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- May 14th 2007, 09:57 PMmyoplex11differential equation problem
solve the following differential equation dy/dx=x[(x^2 + 4)^(1/2)] with initial condition y(0)=1

- May 14th 2007, 11:14 PMCaptainBlack
consider the function:

f(x) = (1/3)(x^2+4)^{3/2}

df/dx = (1/3)(3/2) 2x (x^2+4)^{1/2} = x[(x^2 + 4)^(1/2)]

So you can use this to integrate your DE you have:

y = (1/3)(x^2+4)^{3/2} + C,

when x=0 we require:

1 = (1/3)(4)^{3/2} + C

so:

C = 1-8/3 = -5/3

and so:

y = (1/3)(x^2+4)^{3/2} - 5/3

RonL - May 15th 2007, 04:00 AMSoroban
Hello, myoplex11!

Quote:

Solve: .dy/dx .= .x(x² + 4)^½ .with initial condition y(0) = 1

We have: .dy .= .x(x² + 4)^½ dx

Integrate: .y .= .(1/3)(x² + 4)^3/2 + C

Initial condition: .y(0) = 1

We have: .(1/3)(0 + 4)^3/2 + C .= .1 . . → . . C = -5/3

Therefore: .y .= .(1/3)(x² + 4)^3/2 - (5/3)