1. Solve the differential equation $y'' + 14y' + 49y = 0$ with initial conditions $y(-4) = -1$ and $y'(-4) = 5$. - This is a second-order linear homogeneous differential equation with constant coefficients. - The characteristic equation is $$r^2 + 14r + 49 = 0$$. - Factor or use the quadratic formula: $$(r + 7)^2 = 0$$ so $r = -7$ is a repeated root. - The general solution for repeated roots is $$y = (C_1 + C_2 x)e^{-7x}$$. - Apply initial conditions: - $y(-4) = (C_1 + C_2(-4))e^{28} = -1$ - $y' = (C_2 - 7C_1 - 7C_2 x)e^{-7x}$, so - $y'(-4) = (C_2 - 7C_1 + 28 C_2) e^{28} = 5$ - Simplify: - $e^{28}(C_1 - 4 C_2) = -1$ - $e^{28}(C_2 - 7 C_1 + 28 C_2) = 5$ - $e^{28}(C_1 - 4 C_2) = -1$ - $e^{28}(29 C_2 - 7 C_1) = 5$ - Divide both equations by $e^{28}$: - $C_1 - 4 C_2 = -e^{-28}$ - $29 C_2 - 7 C_1 = 5 e^{-28}$ - Solve the system: - From first: $C_1 = -e^{-28} + 4 C_2$ - Substitute into second: $$29 C_2 - 7(-e^{-28} + 4 C_2) = 5 e^{-28}$$ $$29 C_2 + 7 e^{-28} - 28 C_2 = 5 e^{-28}$$ $$C_2 + 7 e^{-28} = 5 e^{-28}$$ $$C_2 = -2 e^{-28}$$ - Then $C_1 = -e^{-28} + 4(-2 e^{-28}) = -e^{-28} - 8 e^{-28} = -9 e^{-28}$ - Final solution: $$y = (-9 e^{-28} - 2 e^{-28} x) e^{-7x} = e^{-7x - 28}(-9 - 2 x)$$ 2. Solve $4 y'' - 5 y' = 0$ with $y(-2) = 0$ and $y'(-2) = 7$. - Characteristic equation: $$4 r^2 - 5 r = 0$$ - Factor: $$r(4 r - 5) = 0$$ so $r = 0$ or $r = \frac{5}{4}$. - General solution: $$y = C_1 + C_2 e^{\frac{5}{4} x}$$. - Derivative: $$y' = C_2 \frac{5}{4} e^{\frac{5}{4} x}$$. - Apply initial conditions: - $y(-2) = C_1 + C_2 e^{-\frac{5}{2}} = 0$ - $y'(-2) = C_2 \frac{5}{4} e^{-\frac{5}{2}} = 7$ - From second: $$C_2 = \frac{7}{\frac{5}{4} e^{-\frac{5}{2}}} = \frac{7 \cdot 4}{5} e^{\frac{5}{2}} = \frac{28}{5} e^{\frac{5}{2}}$$ - Substitute into first: $$C_1 + \frac{28}{5} e^{\frac{5}{2}} e^{-\frac{5}{2}} = C_1 + \frac{28}{5} = 0$$ $$C_1 = -\frac{28}{5}$$ - Final solution: $$y = -\frac{28}{5} + \frac{28}{5} e^{\frac{5}{4} x}$$ 3. Solve $2 y^{(4)} + 11 y^{(3)} + 18 y'' + 4 y' - 8 y = 0$. - Characteristic equation: $$2 r^4 + 11 r^3 + 18 r^2 + 4 r - 8 = 0$$ - Try rational roots using factors of 8 and 2: possible roots $\pm1, \pm2, \pm4, \pm\frac{1}{2}, \pm\frac{3}{2}$ etc. - Test $r=1$: $$2 + 11 + 18 + 4 - 8 = 27 \neq 0$$ - Test $r=-1$: $$2 - 11 + 18 - 4 - 8 = -3 \neq 0$$ - Test $r=2$: $$2(16) + 11(8) + 18(4) + 4(2) - 8 = 32 + 88 + 72 + 8 - 8 = 192 \neq 0$$ - Test $r=-2$: $$2(16) - 11(8) + 18(4) - 8 - 8 = 32 - 88 + 72 - 8 - 8 = 0$$ - So $r = -2$ is a root. - Divide polynomial by $(r + 2)$: $$2 r^4 + 11 r^3 + 18 r^2 + 4 r - 8 = (r + 2)(2 r^3 + 7 r^2 + 4 r - 4)$$ - Try to factor cubic $2 r^3 + 7 r^2 + 4 r - 4$. - Test $r=1$: $$2 + 7 + 4 - 4 = 9 \neq 0$$ - Test $r=-1$: $$-2 + 7 - 4 - 4 = -3 \neq 0$$ - Test $r=2$: $$16 + 28 + 8 - 4 = 48 \neq 0$$ - Test $r=-2$: $$-16 + 28 - 8 - 4 = 0$$ - So $r = -2$ is again a root. - Divide cubic by $(r + 2)$: $$2 r^3 + 7 r^2 + 4 r - 4 = (r + 2)(2 r^2 + 3 r - 2)$$ - Factor quadratic: $$2 r^2 + 3 r - 2 = (2 r - 1)(r + 2)$$ - So full factorization: $$(r + 2)^3 (2 r - 1) = 0$$ - Roots: $r = -2$ (triple root), $r = \frac{1}{2}$. - General solution: $$y = (C_1 + C_2 x + C_3 x^2) e^{-2 x} + C_4 e^{\frac{x}{2}}$$ 4. Solve the differential equation $(x - 2 y) dx + x dy = 0$ using the homogeneous DE method. - Rewrite as: $$ (x - 2 y) + x \frac{dy}{dx} = 0$$ $$x \frac{dy}{dx} = 2 y - x$$ $$\frac{dy}{dx} = \frac{2 y - x}{x} = 2 \frac{y}{x} - 1$$ - Substitute $v = \frac{y}{x}$, so $y = v x$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$. - Substitute into DE: $$v + x \frac{dv}{dx} = 2 v - 1$$ $$x \frac{dv}{dx} = v - 1$$ $$\frac{dv}{dx} = \frac{v - 1}{x}$$ - Separate variables: $$\frac{dv}{v - 1} = \frac{dx}{x}$$ - Integrate both sides: $$\int \frac{1}{v - 1} dv = \int \frac{1}{x} dx$$ $$\ln |v - 1| = \ln |x| + C$$ - Exponentiate: $$|v - 1| = C x$$ - Substitute back $v = \frac{y}{x}$: $$\left| \frac{y}{x} - 1 \right| = C x$$ $$|y - x| = C x^2$$ - General implicit solution: $$y - x = K x^2$$ where $K = \pm C$. Final answers: 1. $$y = e^{-7 x - 28} (-9 - 2 x)$$ 2. $$y = -\frac{28}{5} + \frac{28}{5} e^{\frac{5}{4} x}$$ 3. $$y = (C_1 + C_2 x + C_3 x^2) e^{-2 x} + C_4 e^{\frac{x}{2}}$$ 4. $$y - x = K x^2$$