Question

Verify that the function satisfies the differential equation. $$ y=\frac{1}{x}, x>0 \quad x^{3} y^{\prime \prime}+2 x^{2} y^{\prime}=0 $$

Short answer

Yes, the function \(y = \frac{1}{x}\) satisfies the provided differential equation, as the substituted and simplified equation resulted in an identity \(0=0\).

Step-by-step solution

Differentiate the function

To start, the function \(y=\frac{1}{x}\) needs to be differentiated. Using the power rule, the first derivative \(y'\) equals to \(-\frac{1}{x^2}\). Then differentiate again to find the second derivative \(y''\). The result is \(y''=2\frac{1}{x^3}\).

Substitute into the equation

Next, substitute \(y'\) and \(y''\) into the equation \(x^{3} y^{\prime \prime}+2 x^{2} y^{\prime}=0\). Replace \(y''\) with \(2\frac{1}{x^3}\) and \(y'\) with \(-\frac{1}{x^2}\) in the equation.

Simplification

After substitution, the equation is \(x^{3} * (2\frac{1}{x^3}) +2 x^{2}*(-\frac{1}{x^2})\). Simplifying the expressions on the right-hand side we get \(2 - 2 = 0\). Which results in \(0=0\). This confirms the function is a solution to the differential equation.

Conclusion

Since the substitution and simplification of the derivatives of the function into the equation produced an identity, this verifies that \(y = \frac{1}{x}\) does indeed satisfy the provided differential equation.