(y-x)y'=y-x+8 Y=x+4(x+2)^(1/2)

Jasonbradley

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Jun 17, 2024

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Solving the Differential Equation (y-x)y' = y-x+8 with the given function y=x+4(x+2)^(1/2)

This article explores the solution to the differential equation (y-x)y' = y-x+8 with the given function y=x+4(x+2)^(1/2). We will verify if the given function is indeed a solution to the differential equation.

1. Finding y'

Firstly, we need to find the derivative of the given function y=x+4(x+2)^(1/2).

Using the power rule and chain rule, we get: y' = 1 + 2(x+2)^(-1/2)

2. Substituting y and y' into the Differential Equation

Now, we substitute y and y' into the original differential equation:

(y-x)y' = y-x+8

[x+4(x+2)^(1/2) - x] * [1 + 2(x+2)^(-1/2)] = [x+4(x+2)^(1/2) - x] + 8

Simplifying the equation:

4(x+2)^(1/2) * [1 + 2(x+2)^(-1/2)] = 4(x+2)^(1/2) + 8

4(x+2)^(1/2) + 8 = 4(x+2)^(1/2) + 8

3. Conclusion

As we can see, both sides of the equation are equal. This verifies that the given function y=x+4(x+2)^(1/2) is indeed a solution to the differential equation (y-x)y' = y-x+8.

In conclusion, we have successfully verified that the given function satisfies the given differential equation.