Verify that the following functions are solutions to the given differential equation. 8. \( y=\frac{x^{3}}{3} \) solves \( y^{\prime}=x^{2} \) 9. \( y=2 e^{-x}+x-1 \) solves \( y^{\prime}=x-y \) 10. \( y=e^{3 x}-\frac{e^{x}}{2} \) solves \( y^{\prime}=3 y+e^{x} \) 11. \( y=\frac{1}{1-x} \) solves \( y^{\prime}=y^{2} \) 12. \( y=e^{x^{2} / 2} \) solves \( y^{\prime}=x y \) 13. \( y=4+\ln x \) solves \( x y^{\prime}=1 \) 14. \( y=3-x+x \ln x \) solves \( y^{\prime}=\ln x \) 15. \( y=2 e^{x}-x-1 \) solves \( y^{\prime}=y+x \)

To verify that each function is a solution to its corresponding differential equation, you'll need to compute the derivative of the given function and check if it satisfies the equation. 1. For \( y = \frac{x^{3}}{3} \): - Calculate \( y' = x^{2} \). The original equation is \( y' = x^{2} \), which holds true. 2. For \( y = 2 e^{-x} + x - 1 \): - Calculate \( y' = -2 e^{-x} + 1 \). Substituting into the original equation \( y' = x - y \) shows that \( -2 e^{-x} + 1 = x - (2 e^{-x} + x - 1) \), confirming it holds. 3. For \( y = e^{3x} - \frac{e^{x}}{2} \): - Calculate \( y' = 3 e^{3x} - \frac{1}{2} e^{x} \). Substituting into \( y' = 3y + e^{x} \) shows the equality holds. 4. For \( y = \frac{1}{1-x} \): - Calculate \( y' = \frac{1}{(1-x)^{2}} \). Substituting shows that \( y' = y^{2} \) holds. 5. For \( y = e^{x^{2}/2} \): - Calculate \( y' = x e^{x^{2}/2} \). This matches the equation \( y' = xy \). 6. For \( y = 4 + \ln x \): - Calculate \( y' = \frac{1}{x} \). The equation \( x y' = 1 \) holds true. 7. For \( y = 3 - x + x \ln x \): - Calculate \( y' = \ln x \). The original equation \( y' = \ln x \) also holds. 8. For \( y = 2 e^{x} - x - 1 \): - Calculate \( y' = 2 e^{x} - 1 \). The original equation \( y' = y + x \) holds true as well. Check each of these derivatives and substitute them back into their respective differential equations to confirm they satisfy the relationships established. Happy differentiating!