# Verify if it's true!y=x-(1/x) is a solution of the differential equation x*y' + y=2x.

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y=x-(1/x) is asolution of:

x*y'+y=2x

==> y= 2x-x*y'

First let us find y'

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

Now: x*y'+y= 2x

==> x(1+1/x^2) + 1-1/x =2x

==> x+ 1/x +1 - 1/x =2x

==> x+1 =2x

==> x=1

We notice that we did not get y value for the function, that means that y value is 0, then y is a solution for the equation.

First, we'll differentiate the given expression y=x-(1/x).

For the ratio (1/x), we'll use the quotient rule.

y' = (x)' - (1/x)'

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

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

Now, we'll extract y' from the given relation x*y' + y=2x.

We'll add -y both sides of the expression.

x*y' + y-y = 2x-y

x*y' = 2x-y

We'll divide by x:

y' = ( 2x-y)/x

Now, we'll substitute y by it's known expression.

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

y' = [x+(1/x)]/x

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

y' = x^2/x^2 + 1/x^2

y' = 1 + 1/x^2

It is obvious that we've obtained the same expressions for y', so the fact that y is a solution of the differential equation is true.

y = x-1/x. To verify whether this is a solution of x*y' +y=2x.

Solution:

y = x-1/x.Multiplying by x both sides, we get:

x*y= x^2 - 1. Differentiating with respect to x, we get:

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

1*y+x*y' = 2x- 0, as d/dx(U*V) = U'V+U*V' and d/dx (kx^n) = knx^(n-1) and d/dx (constant) = 0.

x*y' +y = 2x. So if x*y = x-1/x is asolution of x*y'+y = 2x is satisfied.