Verify that the function y(x)= x - (1/x) is a solution to the differential equation, xy'+ y= 2x
I have absolutely no clue what to do with that + y
Please help :c
y = x - 1/x
y' = 1 + 1/x^2
So, to verify that it is a solution, just plug and chug:
xy' + y = x(1 + 1/x^2) + x - 1/x = x + 1/x + x - 1/x = 2x
To verify if the function y(x) = x - (1/x) is a solution to the differential equation xy' + y = 2x, we need to substitute y(x) and its derivative y'(x) into the equation and check if it satisfies the equation.
First, let's find the derivative of y(x):
y(x) = x - (1/x)
Taking the derivative of both sides with respect to x:
y'(x) = 1 - (-1/x^2)
= 1 + (1/x^2)
Now, substitute y(x) and y'(x) back into the given differential equation:
xy' + y = 2x
x(1 + (1/x^2)) + (x - (1/x)) = 2x
Simplify the equation:
x + (x/x^2) + x - (1/x) = 2x
Multiply each term by x:
x^2 + 1 + x^2 - 1 = 2x^2
2x^2 = 2x^2
Since the equation is satisfied, the function y(x) = x - (1/x) is indeed a solution to the differential equation xy' + y = 2x.
To verify that the function y(x) = x - (1/x) is a solution to the differential equation xy' + y = 2x, we need to substitute this function into the differential equation and see if it satisfies the equation.
1. Start by finding the derivative of y(x) using the power rule and the quotient rule.
y'(x) = 1 - (-1/x^2) = 1 + (1/x^2) = 1/x^2 + 1
2. Substitute the function y(x) and its derivative y'(x) back into the differential equation xy' + y = 2x.
x(1/x^2 + 1) + (x - 1/x) = 2x
Simplify the equation:
1/x + x + x - 1/x = 2x
Combine like terms:
2x = 2x
The equation is satisfied.
Therefore, y(x) = x - (1/x) is indeed a solution to the given differential equation.