Find the solution?

Question

Find the solution of y'+2xy=x, with y(0)=-2.

Answer 1

This is a separable equation- meaning the  'y' terms and 'x' terms can be separated. First, we bring the 2xy term to the right side of the equation:

y' =x-2xy, which simplifies to

y' = x(1-2y). dividing by 1-2y, we get

y'/(1-2y) = x which we then rewrite as

dy/(1-2y) = xdx. Now, we integrate both sides.

The right side is simply 1/2x² + c. Don't forget the + c

The left side we should, by now, recognize as  a log function. By saying U= 1-2y, dU = -2dy, so dy= -dU/2

Thus, we are integrating -dU/2U, which is just -log(U)/2. Substituting back in U= 1-2y, we get

-log(1-2y)/2 +c₁ = 1/2x² +c_2. Combining the two constants (since they are just arbitrary constants), and multiplying by -2 for simplicity, we get

log(1-2y) = -x² +c. Raising e to the power of both sides, we find

1-2y= e^{-x^2 +c}, which is the same as 

1-2y= e^ce^{-x^2}. Since c is just an arbitrary constant, we can just replace e^c with c. Thus 

1-2y= Ce^{-x^2}. Rearranging the terms, we get

y= (1-Ce^{-x^2})/2. Since we know y(0)= -2, we get (1-C)/2= -2. This means C=5. Our final solution is then

y(x)= (1-5e^{-x^2})/2

 

Answer 2

You can use separtion of variables to solve the problem. But this is a typical first-order linear differential equation: y' + Py = Q

Here is a standard way to solve it.

Multiply both sides of y' + Py = Q by e^∫P,

e^∫P y' + e^∫P Py = e^∫PQ

Apply product rule,

(e^∫P y)' = e^∫P Q

Solve for y,

y = (1/e^∫P)(∫e^∫P Q dx + c) <==Formula of the solution for any first-order linear differential equation

For your question,

P = 2x

e^∫P = e^{x^2}

y = e^{-x^2} (∫e^{x^2} x dx + c)
 = e^{-x^2} [(1/2)e^{x^2} + c]

Plug in y(0) = -2,

-2 = (1/2) + c => c = -5/2)

Answer: y = (1/2)[1 - 5e^{-x^2}]

 

Answer 3

The key to solving this problem is to use the product rule in calculus, namely that (xy)' = xy' + x'y. To get this, multiply the entire equation by z:

zy' + 2xyz = xz.

The crux move is figuring out what z is by the product rule:

zy' + 2xyz = zy' + yz';

2xyz = yz';

z' = 2xz.

Now, what function's derivative would be a real number multiplied by itself? Obviously, it's a function in the e^a family:

(e^a)' = a'e^a.

With that in mind, it is not difficult to figure out that a should be x², so z = e^{x^2}. Now, the function looks like this:

e^{x^2}y' + 2xye^{x^2} = xe^{x^2}.

We can now integrate both sides of the equation. Since we set up the left side so that the product rule would work, the function looks like this:

ye^{x^2} = integrate(xe^{x^2} dx)

        = 1/2e^{x^2} + C;

Now divide the function by e^{x^2}:

y       = 1/2 + Ce^{-x^2}.

We know from the initial condition that y(0) = -2, so plug 0 in for x into the equation and -2 for y:

-2 = 1/2 + C;

C  = -5/2.

Hence, the final answer to the question is y = 1/2 - (5/2)e^{-x^2}.

I believe that this is the solution method that Boyce and Diprima recommended for you?