Verify if it's true!y=x-(1/x) is a solution of the differential equation x*y' + y=2x.
y=x-(1/x) is asolution of:
==> y= 2x-x*y'
First let us find y'
Now: x*y'+y= 2x
==> x(1+1/x^2) + 1-1/x =2x
==> x+ 1/x +1 - 1/x =2x
==> x+1 =2x
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.
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.