Thread: differential equation problem

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

Originally Posted by myoplex11

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

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

Hello, myoplex11!

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)