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)