1 Answer | Add Yours
You need to keep the terms in y to one side and the terms in x to the other side, hence, you need to multiply both sides by dx such that:
`(dy*dx)/(dx) = 16x*e^4x*dx`
Reducing by dx yields:
`dy = 16x*e^4x*dx`
You need to use the reverse of differentiation to find y, hence you need to integrate both sides such that:
`int dy = int 16x*e^4x*dx`
Taking 16 out yields
`int dy = 16int x*e^(4x)*dx`
You should apply integration by parts to solve the right integral. You need to remember the formula of integration by parts.
`int udv = uv - int vdu`
You should consider u = x because differentiating, the power of x decreases.
u=x => du = dx
`dv = e^(4x)dx =gt v = (e^(4x))/4`
`int x*e^(4x)*dx = (x*e^(4x))/4 - int (e^(4x)dx)/4`
Multiplying by 16 both sides, yields:
`16int x*e^(4x)*dx = 16(x*e^(4x))/4 - 16(e^(4x)dx)/16 + c`
`16int x*e^(4x)*dx = 4(x*e^(4x)) - e^(4x) + c`
Factoring out `e^(4x)` yields:
`16int x*e^(4x)*dx = e^(4x)*(4x - 1) + c`
Hence, `int dy = 16int x*e^(4x)*dx =gt y = e^(4x)*(4x - 1) + c`
You need to find the constant c, hence you need to use the information y(0)=3.
Plugging x=0 in `y = e^(4x)*(4x - 1) + c` yields:
`y(0) = e^(4*0)*(4*0 - 1) + c =gt y(0) = e^0*(-1) + c`
`y(0) = 1*(-1) + c =gt 3 = -1 + c =gt c = 3 + 1 =gt c = 4`
Hence, evaluating the particular solution of ordinary differential equation yields `y = e^(4x)*(4x - 1) + 4` .
We’ve answered 319,199 questions. We can answer yours, too.Ask a question