We want to solve the Initial Value Problem y' = y + 4xy, with y(0) = 1.
To use Euler's method, define
y(i+1) = y(i) + hy'(i), for i=0,1,2, ...,
where
h = 0.1, the step size.,
x(i) = i*h
1st step.
y(0) = 1 (given) and x(0) = 0.
y(1) ≡ y(0.1) = y(0) + h*[4*x(0)*y(0)] = 1
2nd step.
x(1) = 0.1
y(2) ≡ y(0.2) = y(1) + h*[4*x(1)*y(1)] = 1 + 0.1*(4*0.1*1) = 1.04
3rd step.
x(2) = 0.2
y(3) ≡ y(0.3) = y(2) + h*[4*x(2)*y(2)] = 1.04 + 0.1*(4*0.2*1.04) = 1.1232
4th step.
x(3) = 0.3
y(4) ≡ y(0.4) = y(3) + h*[4*x(3)*y(3)] = 1.1232 + 0.1*(4*0.3*1.1232) = 1.258
5th step.
x(4) = 0.4
y(5) ≡ y(0.5) = y(4) + h*[4*x(4)*y(4)] = 1.258 + 0.1*(4*0.4*1.258) = 1.4593
Answer: y(0.5) = 1.4593
I'll assume you're supposed to compute the line integral of over the given path . By the fundamental theorem of calculus,
so evaluating the integral is as simple as evaluting at the endpoints of . But first we need to determine given its gradient.
We have
Differentiating with respect to gives
and we end up with
for some constant . Then the value of the line integral is .
3t - 15 = 2t + 10
t = 25
<Y = 2(25) + 10 = 60
<X = 4r
so
4r + 60 = 180
4r = 120 (<X = 4r)
answer
<X = 120
Combine like terms on #13 and that's all you can do on that one because there's nothing that'll go into all three of those. So the answer will just be two x squared minus x minus 36