# Euler's and improved Euler's method,...find the exact solution

1. y'=1/2 -x + 2y y(0) = 1
Find the exact solution "Sphee"
{0 with line diagonal of the 0}

2. y'=x square + y square
y(0)=1

a. Let h=.1 use Euler and improve to approximate to get
Sphee(.1) Sphee(.2) Sphee(.3)
b. h=.05 so compare with Shee(.1)
Sphee (.2) sphee is a circle

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#### Solution Preview

...)=2 f(x)(e^(2x))+1/2-x Thus.

f'(x)=(1/2-x)/ e^(2x)= (1/2-x)e^(-2x) =(1/2) e^(-2x)-xe^(-2x) Now we integrate both sides

Use integration by parts formula, let u=x, v=e^(-2x), dv=-2 e^(-2x)dx

We have the integration of udv= uv-the integration of vdu

we have the integration of xe^(-2x)dx =-(1/2)udv is equal to -(1/2)(uv- integration of vdu).

Thus the integration of xe^(-2x)dx=-(1/2)[x e^(-2x)+(1/2) e^(-2x) ]

Thus

f(x)=(-1/4) e^(-2x) +(1/2)(xe^(-2x)+(1/2) e^(-2x))+C=1/2xe^(-2x)+C

Thus the general solution is

y=f(x)(e^(2x))= (1/2)xe^(-2x)+C) (e^(2x))=(1/2)x+C(e^(2x)). With y(0)=1,

we have C=1, Thus the solution for the exact solution for the given problem is

y=(1/2)x+C(e^(2x))= (1/2)x+(e^(2x)).

Don't understand what you by

Find the exact solution "Sphee"
{0 with line diagnal of the 0}

Do you mean x^2+y^2=0 which implies that x=0, y=0. If not please clarify next time.

If we need to find the exact solution when x=0, y=0

If x=0, y=0. From the ...