1. **Problem:** Show that each given function is a solution of the corresponding differential equation. 2. **General approach:** To verify a solution, substitute the function and its derivatives into the differential equation and simplify to check if the equation holds true. --- **1.** Given $y = x^2 + cx$, verify $xy' = x^2 + y$. - Compute $y' = 2x + c$. - Substitute into equation: $x(2x + c) = x^2 + (x^2 + cx)$. - Left side: $2x^2 + cx$. - Right side: $x^2 + x^2 + cx = 2x^2 + cx$. - Both sides equal, so solution verified. **2.** Given $y = \frac{c}{x} + 1$, verify $xy' + y = 1$. - 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. **3.** Given $y = 3 \tan 3x$, verify $y' = 9 + y^2$. - Compute $y' = 3 \cdot 3 \sec^2 3x = 9 \sec^2 3x$. - Recall $\sec^2 3x = 1 + \tan^2 3x$. - Substitute $y$: $y' = 9(1 + (\frac{y}{3})^2) = 9 + y^2$. - Verified. **4.** Given $y = e^{-x} + \sin x$, verify $y'' + y = 2e^{-x}$. - Compute $y' = -e^{-x} + \cos x$. - Compute $y'' = e^{-x} - \sin x$. - Substitute: $y'' + y = (e^{-x} - \sin x) + (e^{-x} + \sin x) = 2e^{-x}$. - Verified. **5.** Given $y = Ae^x + Be^{-2x} + x^2 + x$, verify $y'' + y' - 2y = 3 - 2x^2$. - 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: $$= Ae^x + 4Be^{-2x} + 2 + Ae^x - 2Be^{-2x} + 2x + 1 - 2Ae^x - 2Be^{-2x} - 2x^2 - 2x$$ - Combine like terms: $$= (Ae^x + Ae^x - 2Ae^x) + (4Be^{-2x} - 2Be^{-2x} - 2Be^{-2x}) + (2 + 1) + (2x - 2x) - 2x^2$$ $$= 0 + 0 + 3 + 0 - 2x^2 = 3 - 2x^2$$ - Verified. **6.** Given $y = e^{3x} \cos 2x$, verify $y'' - 6y' + 13y = 0$. - 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: $$y'' - 6y' + 13y = e^{3x}(5 \cos 2x - 12 \sin 2x) - 6 e^{3x}(3 \cos 2x - 2 \sin 2x) + 13 e^{3x} \cos 2x$$ - Simplify: $$= e^{3x}[5 \cos 2x - 12 \sin 2x - 18 \cos 2x + 12 \sin 2x + 13 \cos 2x] = e^{3x}[0] = 0$$ - Verified. **7.** Given implicit $y^3 - 3x + 3y = 5$, verify $y'' + 2y(y')^3 = 0$. - 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 $y'$: $$y'' = \frac{d}{dx} \left( \frac{1}{y^2 + 1} \right) = - \frac{2y y'}{(y^2 + 1)^2}$$ - 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: $$y'' + 2y(y')^3 = - \frac{2y}{(y^2 + 1)^3} + \frac{2y}{(y^2 + 1)^3} = 0$$ - Verified. **8.** Given $\log y + \frac{x}{y} = c$, verify $(y - x) y' + y = 0$. - Differentiate implicitly: $$\frac{y'}{y} + \frac{y - x y'}{y^2} = 0$$ - Multiply both sides by $y^2$: $$y y' + y - x y' = 0$$ - Rearrange: $$(y - x) y' + y = 0$$ - Verified. **9.** Given parametric $x = 4t + 1$, $y = t^2 - 2$, verify $4(y')^2 = y + 2$. - Compute derivatives: $$\frac{dx}{dt} = 4, \quad \frac{dy}{dt} = 2t$$ - Compute $y' = \frac{dy/dt}{dx/dt} = \frac{2t}{4} = \frac{t}{2}$. - Compute $4(y')^2 = 4 \left( \frac{t}{2} \right)^2 = 4 \cdot \frac{t^2}{4} = t^2$. - Compute $y + 2 = (t^2 - 2) + 2 = t^2$. - Both sides equal, verified. --- All functions satisfy their respective differential equations as required.