1. **Find the value of $k$ for which $y = e^{kx}$ is a solution to the differential equation** Given differential equation: $$y'' - 7y' + 14y' - 8y = 0$$ First, simplify the equation: $$y'' + ( -7 + 14 ) y' - 8y = 0 \implies y'' + 7y' - 8y = 0$$ Assume solution $y = e^{kx}$, then: $$y' = ke^{kx}, \quad y'' = k^2 e^{kx}$$ Substitute into the differential equation: $$k^2 e^{kx} + 7k e^{kx} - 8 e^{kx} = 0$$ Divide both sides by $e^{kx}$ (never zero): $$k^2 + 7k - 8 = 0$$ Solve quadratic: $$k = \frac{-7 \pm \sqrt{7^2 - 4 \times 1 \times (-8)}}{2} = \frac{-7 \pm \sqrt{49 + 32}}{2} = \frac{-7 \pm \sqrt{81}}{2}$$ $$k = \frac{-7 \pm 9}{2}$$ So, $$k_1 = \frac{-7 + 9}{2} = 1, \quad k_2 = \frac{-7 - 9}{2} = -8$$ **Answer:** $k = 1$ or $k = -8$ --- 2. **Find the order, degree and linearity of the differential equations:** (i) $$\left(\frac{d^2 y}{dx^2}\right)^{\frac{3}{2}} = \frac{dy}{dx} + y + 2$$ - Order: Highest derivative is $\frac{d^2 y}{dx^2}$, so order = 2. - Degree: The highest power of the highest order derivative after removing radicals and fractions is found by rewriting: $$\left(\frac{d^2 y}{dx^2}\right)^{\frac{3}{2}} = \left(\frac{d^2 y}{dx^2}\right)^{1.5}$$ Since the derivative is raised to a fractional power, degree is not defined (degree must be a positive integer). - Linearity: The equation is nonlinear because the highest order derivative is raised to a power other than 1. (ii) $$\sqrt{3 + \left(\frac{dy}{dx}\right)^2} = \sec x$$ - Order: Highest derivative is $\frac{dy}{dx}$, so order = 1. - Degree: The derivative appears inside a square root, so degree is not defined. - Linearity: Nonlinear because of the square root and the derivative squared. --- 3. **Solve the initial value problem:** $$x \frac{dy}{dx} + \cot y = 0, \quad y(\sqrt{2}) = \frac{\pi}{4}$$ Rewrite: $$x \frac{dy}{dx} = -\cot y$$ $$\frac{dy}{dx} = -\frac{\cot y}{x}$$ Separate variables: $$\frac{dy}{\cot y} = -\frac{dx}{x}$$ Recall $\cot y = \frac{\cos y}{\sin y}$, so: $$\frac{dy}{\cot y} = \frac{dy}{\frac{\cos y}{\sin y}} = \frac{\sin y}{\cos y} dy = \tan y \, dy$$ So: $$\tan y \, dy = -\frac{dx}{x}$$ Integrate both sides: $$\int \tan y \, dy = - \int \frac{dx}{x}$$ Recall: $$\int \tan y \, dy = -\ln |\cos y| + C$$ So: $$-\ln |\cos y| = -\ln |x| + C$$ Multiply both sides by -1: $$\ln |\cos y| = \ln |x| - C$$ Let $C_1 = e^C$, then: $$|\cos y| = C_1 |x|$$ Use initial condition $y(\sqrt{2}) = \frac{\pi}{4}$: $$\cos \frac{\pi}{4} = C_1 \sqrt{2} \implies \frac{\sqrt{2}}{2} = C_1 \sqrt{2} \implies C_1 = \frac{1}{2}$$ Final implicit solution: $$|\cos y| = \frac{|x|}{2}$$ --- 4. **Given differential equation:** $$(x^4 - 2xy^2 + y^4) dx - (2x^2 y - 4xy^3 + \sin y) dy = 0$$ (i) Show it is exact. Let: $$M = x^4 - 2xy^2 + y^4, \quad N = -(2x^2 y - 4xy^3 + \sin y) = -2x^2 y + 4xy^3 - \sin y$$ Calculate partial derivatives: $$\frac{\partial M}{\partial y} = -4xy + 4y^3$$ $$\frac{\partial N}{\partial x} = -4xy + 4y^3$$ Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is exact. (ii) Solve the differential equation. Find potential function $\psi(x,y)$ such that: $$\frac{\partial \psi}{\partial x} = M = x^4 - 2xy^2 + y^4$$ Integrate w.r.t. $x$: $$\psi = \int (x^4 - 2xy^2 + y^4) dx = \frac{x^5}{5} - x^2 y^2 + x y^4 + h(y)$$ Differentiate $\psi$ w.r.t. $y$: $$\frac{\partial \psi}{\partial y} = -2x^2 y + 4x y^3 + h'(y)$$ Set equal to $N$: $$-2x^2 y + 4x y^3 + h'(y) = -2x^2 y + 4x y^3 - \sin y$$ So: $$h'(y) = -\sin y$$ Integrate: $$h(y) = \cos y + C$$ General solution: $$\frac{x^5}{5} - x^2 y^2 + x y^4 + \cos y = C$$ --- 5. **Verify homogeneity and solve:** $$(x^2 + xy + y^2) dy - (2xy + y^2) dx = 0$$ Rewrite: $$M = -(2xy + y^2), \quad N = x^2 + xy + y^2$$ Check homogeneity: - Degree of each term in $M$: - $2xy$: degree 2 - $y^2$: degree 2 - Degree of each term in $N$: - $x^2$: degree 2 - $xy$: degree 2 - $y^2$: degree 2 Both $M$ and $N$ are homogeneous functions of degree 2. Substitute $y = vx \Rightarrow dy = v dx + x dv$ Rewrite equation: $$(x^2 + x(vx) + (vx)^2)(v dx + x dv) - (2x(vx) + (vx)^2) dx = 0$$ Simplify: $$(x^2 + v x^2 + v^2 x^2)(v dx + x dv) - (2 v x^2 + v^2 x^2) dx = 0$$ $$x^2 (1 + v + v^2)(v dx + x dv) - x^2 (2 v + v^2) dx = 0$$ Divide by $x^2$: $$(1 + v + v^2)(v dx + x dv) - (2 v + v^2) dx = 0$$ Expand: $$v (1 + v + v^2) dx + x (1 + v + v^2) dv - (2 v + v^2) dx = 0$$ Group $dx$ terms: $$[v (1 + v + v^2) - (2 v + v^2)] dx + x (1 + v + v^2) dv = 0$$ Simplify bracket: $$v + v^2 + v^3 - 2 v - v^2 = v^3 - v$$ So: $$(v^3 - v) dx + x (1 + v + v^2) dv = 0$$ Rewrite: $$ (v^3 - v) dx = - x (1 + v + v^2) dv$$ Divide both sides by $x$: $$\frac{dx}{x} = - \frac{1 + v + v^2}{v^3 - v} dv$$ Integrate both sides: $$\int \frac{dx}{x} = - \int \frac{1 + v + v^2}{v^3 - v} dv$$ Factor denominator: $$v^3 - v = v (v^2 - 1) = v (v - 1)(v + 1)$$ Use partial fractions for the integral on right: $$\frac{1 + v + v^2}{v (v - 1)(v + 1)} = \frac{A}{v} + \frac{B}{v - 1} + \frac{C}{v + 1}$$ Multiply both sides by denominator: $$1 + v + v^2 = A (v - 1)(v + 1) + B v (v + 1) + C v (v - 1)$$ Expand: $$A (v^2 - 1) + B v^2 + B v + C v^2 - C v$$ Group terms: $$(A + B + C) v^2 + (B - C) v - A = 1 + v + v^2$$ Equate coefficients: - $v^2$: $A + B + C = 1$ - $v$: $B - C = 1$ - constant: $-A = 1 \implies A = -1$ Substitute $A = -1$: $$-1 + B + C = 1 \implies B + C = 2$$ $$B - C = 1$$ Add equations: $$2B = 3 \implies B = \frac{3}{2}$$ Then: $$C = 2 - B = 2 - \frac{3}{2} = \frac{1}{2}$$ Integral becomes: $$- \int \left( \frac{-1}{v} + \frac{3/2}{v - 1} + \frac{1/2}{v + 1} \right) dv = - \int \left(-\frac{1}{v} + \frac{3}{2(v - 1)} + \frac{1}{2(v + 1)} \right) dv$$ Integrate termwise: $$- \left(-\ln |v| + \frac{3}{2} \ln |v - 1| + \frac{1}{2} \ln |v + 1| \right) + C$$ $$= \ln |v| - \frac{3}{2} \ln |v - 1| - \frac{1}{2} \ln |v + 1| + C$$ Left side integral: $$\int \frac{dx}{x} = \ln |x| + C$$ Equate: $$\ln |x| = \ln |v| - \frac{3}{2} \ln |v - 1| - \frac{1}{2} \ln |v + 1| + C$$ Rewrite: $$\ln |x| - \ln |v| + \frac{3}{2} \ln |v - 1| + \frac{1}{2} \ln |v + 1| = C$$ Recall $v = \frac{y}{x}$: $$\ln \left| \frac{x}{v} \right| + \frac{3}{2} \ln |v - 1| + \frac{1}{2} \ln |v + 1| = C$$ $$\ln |\frac{x}{y/x}| + \frac{3}{2} \ln |\frac{y}{x} - 1| + \frac{1}{2} \ln |\frac{y}{x} + 1| = C$$ $$\ln |\frac{x^2}{y}| + \frac{3}{2} \ln |\frac{y - x}{x}| + \frac{1}{2} \ln |\frac{y + x}{x}| = C$$ Combine logarithms: $$\ln \left| \frac{x^2}{y} \times \left( \frac{y - x}{x} \right)^{3/2} \times \left( \frac{y + x}{x} \right)^{1/2} \right| = C$$ $$\ln \left| \frac{x^2}{y} \times \frac{(y - x)^{3/2} (y + x)^{1/2}}{x^{3/2 + 1/2}} \right| = C$$ $$\ln \left| \frac{x^2}{y} \times \frac{(y - x)^{3/2} (y + x)^{1/2}}{x^2} \right| = C$$ $$\ln \left| \frac{(y - x)^{3/2} (y + x)^{1/2}}{y} \right| = C$$ Exponentiate both sides: $$\left| \frac{(y - x)^{3/2} (y + x)^{1/2}}{y} \right| = K$$ where $K = e^C$ is constant. **Final implicit solution:** $$\frac{(y - x)^{3/2} (y + x)^{1/2}}{y} = C$$