1. Problem: Verify that each given function satisfies its corresponding differential equation. 2. For each function, we will: - Differentiate as needed. - Substitute into the differential equation. - Simplify and check if the equation holds true. 3. Example 1: $y = x^2 + cx$, equation: $xy' = x^2 + y$ - Compute $y' = 2x + c$ - Substitute: $x(2x + c) = x^2 + (x^2 + cx)$ - Left side: $2x^2 + cx$ - Right side: $x^2 + x^2 + cx = 2x^2 + cx$ - Both sides equal, so solution holds. 4. Example 2: $y = \frac{c}{x} + 1$, equation: $xy' + y = 1$ - Compute $y' = -\frac{c}{x^2}$ - Substitute: $x(-\frac{c}{x^2}) + (\frac{c}{x} + 1) = -\frac{c}{x} + \frac{c}{x} + 1 = 1$ - Equation holds. 5. Example 3: $y = 3 \tan 3x$, equation: $y' = 9 + y^2$ - Compute $y' = 3 \cdot 3 \sec^2 3x = 9 \sec^2 3x$ - Note $\sec^2 3x = 1 + \tan^2 3x$ - Substitute: $y' = 9(1 + \tan^2 3x) = 9 + 9 \tan^2 3x = 9 + y^2$ - Equation holds. 6. Example 4: $y = e^{-x} + \sin x$, equation: $y'' + y = 2e^{-x}$ - Compute $y' = -e^{-x} + \cos x$ - Compute $y'' = e^{-x} - \sin x$ - Substitute: $y'' + y = (e^{-x} - \sin x) + (e^{-x} + \sin x) = 2e^{-x}$ - Equation holds. 7. Example 5: $y = Ae^x + Be^{-2x} + x^2 + x$, equation: $y'' + y' - 2y = 3 - 2x^2$ - Compute $y' = Ae^x - 2Be^{-2x} + 2x + 1$ - Compute $y'' = Ae^x + 4Be^{-2x} + 2$ - Substitute: $$y'' + y' - 2y = (Ae^x + 4Be^{-2x} + 2) + (Ae^x - 2Be^{-2x} + 2x + 1) - 2(Ae^x + Be^{-2x} + x^2 + x)$$ - Simplify: $$= Ae^x + 4Be^{-2x} + 2 + Ae^x - 2Be^{-2x} + 2x + 1 - 2Ae^x - 2Be^{-2x} - 2x^2 - 2x$$ $$= (Ae^x + Ae^x - 2Ae^x) + (4Be^{-2x} - 2Be^{-2x} - 2Be^{-2x}) + (2 + 1) + (2x - 2x) - 2x^2$$ $$= 0 + 0 + 3 + 0 - 2x^2 = 3 - 2x^2$$ - Equation holds. 8. Example 6: $y = e^{3x} \cos 2x$, equation: $y'' - 6y' + 13y = 0$ - Compute $y' = e^{3x}(3 \cos 2x - 2 \sin 2x)$ - Compute $y'' = e^{3x}((3)^2 \cos 2x - 2 \cdot 3 \sin 2x - 2 \cdot 3 \sin 2x - 4 \cos 2x) = e^{3x}(9 \cos 2x - 12 \sin 2x - 4 \cos 2x) = e^{3x}(5 \cos 2x - 12 \sin 2x)$ - Substitute: $$y'' - 6y' + 13y = e^{3x}(5 \cos 2x - 12 \sin 2x) - 6 e^{3x}(3 \cos 2x - 2 \sin 2x) + 13 e^{3x} \cos 2x$$ $$= e^{3x}[5 \cos 2x - 12 \sin 2x - 18 \cos 2x + 12 \sin 2x + 13 \cos 2x] = e^{3x}[0] = 0$$ - Equation holds. 9. Example 7: Implicit function $y^3 - 3x + 3y = 5$, equation: $y'' + 2y(y')^3 = 0$ - Differentiate implicitly: $$3y^2 y' - 3 + 3 y' = 0 \Rightarrow y'(3y^2 + 3) = 3 \Rightarrow y' = \frac{3}{3(y^2 + 1)} = \frac{1}{y^2 + 1}$$ - Differentiate $y'$ again: $$y'' = \frac{d}{dx} \left( \frac{1}{y^2 + 1} \right) = - \frac{2y y'}{(y^2 + 1)^2}$$ - Substitute $y'$: $$y'' = - \frac{2y \cdot \frac{1}{y^2 + 1}}{(y^2 + 1)^2} = - \frac{2y}{(y^2 + 1)^3}$$ - Compute $2y(y')^3 = 2y \left( \frac{1}{y^2 + 1} \right)^3 = \frac{2y}{(y^2 + 1)^3}$ - Sum: $$y'' + 2y(y')^3 = - \frac{2y}{(y^2 + 1)^3} + \frac{2y}{(y^2 + 1)^3} = 0$$ - Equation holds. 10. Example 8: $\log y + \frac{x}{y} = c$, equation: $(y - x) y' + y = 0$ - Differentiate implicitly: $$\frac{y'}{y} + \frac{y - x y'}{y^2} = 0 \Rightarrow \frac{y'}{y} + \frac{y}{y^2} - \frac{x y'}{y^2} = 0$$ $$\Rightarrow \frac{y'}{y} + \frac{1}{y} - \frac{x y'}{y^2} = 0$$ Multiply both sides by $y^2$: $$y y' + y - x y' = 0 \Rightarrow (y - x) y' + y = 0$$ - Equation holds. 11. Example 9: Parametric $x = 4t + 1$, $y = t^2 - 2$, equation: $4(y')^2 = y + 2$ - Compute $y' = \frac{dy/dt}{dx/dt} = \frac{2t}{4} = \frac{t}{2}$ - Substitute: $$4 \left( \frac{t}{2} \right)^2 = 4 \cdot \frac{t^2}{4} = t^2$$ $$y + 2 = (t^2 - 2) + 2 = t^2$$ - Equation holds. Final: All given functions satisfy their respective differential equations as required.