find the complete solution.(1 + x raise 2) y’ + xy = x raise -1 (1 + x raise 2)raise 1/2?

Answer 1

If you have

y' + f(x) y = g(x)

then let F(x) be such that F'(x) = f(x)

Multiply both sides by e^(F(x)) to get

e^(F(x)) (y' + f(x) y) = g(x) e^(F(x))

Then notice that

(y e^(F(x))' = y' e^(F(x)) + y F'(x) e^(F(x)) = e^(F(x)) (y' + F'(x) y)
= e^F(x) (y' + f(x) y) = left hand side!

So

(y e^(F(x)))' = g(x) e^(F(x))

y e^F(x) = integral of g(x) e^(F(x))

y = e^(- F(x)) integral of g(x) e^(F(x))

Your equation is

Let me rewrite that as:

(1 + x^2) y’ + xy = x^(-1) (1 + x^2)^(1/2) = sqrt(1 + x^2) / x

which after division by (1 + x^2) has this form

y' + (x / (1 + x^2)) y = 1 / (x sqrt(1 + x^2))

So f(x) = x / (1 + x^2), g(x) = 1 / (x sqrt(1 + x^2)), and F(x) is such that F'(x) = f(x) so F(x) = (1/2) ln(1 + x^2) will do for F(x).

Then you have e^(F(x)) = sqrt(1 + x^2)

So the solution is

y(x) = (1 / sqrt(1 + x^2)) integral of g(x) sqrt(1 + x^2)
= (1 / sqrt(1 + x^2)) integral of 1 / x
= (1 / sqrt(1 + x^2)) (ln(x) + C)

I'll let you verify that

(1 + x^2) y’ + xy = x^(-1) (1 + x^2)^(1/2) = sqrt(1 + x^2) / x

Answer 2

Divide by (1+x^2) to get the y' term by itself, then simplify the right side:

y' + x/(1+x^2)*y = x^-1 * (1+x^2)^(1/2) / (1+x^2)
y' + x/(1+x^2)*y = (x^-1)*(1+x^2)^(-1/2)

Solve this problem by multiply through by the integrating factor, which is:

I(x)=e^int(x/1+x^2)dx

first you have to use a u sub to solve this one.
Let u=1+x^2
du=2x*dx
dx=du/2x

I(x)=e^(1/2)int(1/u)du= e^(1/2*ln(u))=e^(1/2*ln(1+x^2))=e^ln(1+x^2)^1/2=(1+x^2)^1/2

(1+x^2)^1/2*[y' + x/(1+x^2)*y = (x^-1)*(1+x^2)^(-1/2)]

(1+x^2)^1/2*y' + (1+x^2)^1/2*x/(1+x^2)*y = (1+x^2)^1/2*(x^-1)*(1+x^2)^(-1/2)

Simplify a bit:

(1+x^2)^1/2*y' + x*(1+x^2)^(-1/2)*y = (1+x^2)*(x^-1)

The reason you use the integrating factor is to shape the left hand side into a chain rule, but in reverse, as to compact two terms into a differential form to integrate:

d/dx[y*(1+x^2)^(1/2)]=(1+x^2)/(x)

Integrate wrt x:

y*(1+x^2)^(1/2)=int[(1+x^2)/(x)]

y*(1+x^2)^(1/2)=int[(1/x+x)]

y*(1+x^2)^(1/2)=ln(x)+(1/2)x^2+C

Solve for y:

y=[ln(x)+(1/2)x^2+C]/[(1+x^2)^(1/2)]

Use your initial conditions to solve for the constant of integration, C.

Done...

Answer 3

Divide everything by 1 + x^2
You now have the following:
y' + (x/(1+x^2))y = 1/[xsqrt(1+x^2)]
Your integrating factor is x/(1+x^2)
The integral of x/(1+x^2) is (1/2)ln(1+x^2)
e^[(1/2)ln(1+x^2)]
= (1+x^2)^(1/2)
= sqrt(1+x^2)
Multiply both sides by sqrt(1+x^2)
sqrt(1+x^2)y' + sqrt(1+x^2)(x/(1+x^2))y = sqrt(1+x^2)1/[xsqrt(1+x^2)]
(d/dx)[sqrt(1+x^2)y] = 1/x
(d/dx)[sqrt(1+x^2)y] = 1/x
Integrate both sides.
sqrt(1+x^2)y = integral [1/x]
sqrt(1+x^2)y = ln(x) + C
Divide everything by sqrt(1+x^2)
y = ln(x)/sqrt(1+x^2) + C/sqrt(1+x^2)

Answer 4

Try Google

Answer 5

im 1,000,000 % sure theres a number in the answer.

Answer 6

Do your own homework!

Answer 7

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