1. **Problem Statement:** Explain the terms order, degree, linear differential equation, and analyze the given differential equations. 2. **Order of a Differential Equation:** The order is the highest derivative present in the equation. For example, if the highest derivative is $\frac{d^3y}{dx^3}$, the order is 3. 3. **Degree of a Differential Equation:** The degree is the power of the highest order derivative after the equation is free from radicals and fractions in derivatives. For example, if the highest derivative is raised to the power 32, the degree is 32. 4. **Linear Differential Equation:** A differential equation is linear if the dependent variable and its derivatives appear to the power 1 and are not multiplied together. The general form is $a_n(x)\frac{d^ny}{dx^n} + \cdots + a_1(x)\frac{dy}{dx} + a_0(x)y = g(x)$. 5. **Analyze the equation:** $$\left(\frac{d^3y}{dx^3}\right)^{32} + \left(\frac{d^2y}{dx^2}\right)^{11} + \left(\frac{dy}{dx}\right)^{20} + y = x$$ - Highest derivative is $\frac{d^3y}{dx^3}$ so order = 3. - Degree is 32 (power of highest derivative). - Since derivatives are raised to powers other than 1, the equation is nonlinear. 6. **Show the equation is exact:** $$3x(xy - 2)dx + (x^3 + 2y)dy = 0$$ - Let $M = 3x(xy - 2) = 3x^2y - 6x$, $N = x^3 + 2y$ - Compute partial derivatives: $$\frac{\partial M}{\partial y} = 3x^2$$ $$\frac{\partial N}{\partial x} = 3x^2$$ - Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is exact. 7. **General solution of:** $$4y'' - 12y' + 5y = 0$$ - Characteristic equation: $$4r^2 - 12r + 5 = 0$$ - Solve using quadratic formula: $$r = \frac{12 \pm \sqrt{(-12)^2 - 4 \times 4 \times 5}}{2 \times 4} = \frac{12 \pm \sqrt{144 - 80}}{8} = \frac{12 \pm \sqrt{64}}{8}$$ $$r = \frac{12 \pm 8}{8}$$ - Roots: $r_1 = \frac{20}{8} = 2.5$, $r_2 = \frac{4}{8} = 0.5$ - General solution: $$y = C_1 e^{2.5x} + C_2 e^{0.5x}$$ 8. **Laplace transform of $f(x) = 1$:** - Definition: $$\mathcal{L}\{1\} = \int_0^{\infty} e^{-sx} \cdot 1 \, dx = \int_0^{\infty} e^{-sx} dx$$ - Evaluate integral: $$= \left[-\frac{1}{s} e^{-sx} \right]_0^{\infty} = 0 + \frac{1}{s} = \frac{1}{s}$$ 9. **Differential equation of family of circles:** - Given: $$\frac{dy}{dx} + \frac{x}{y} = 0$$ - Rearrange: $$\frac{dy}{dx} = -\frac{x}{y}$$ - Separate variables: $$y dy = -x dx$$ - Integrate both sides: $$\int y dy = - \int x dx$$ $$\frac{y^2}{2} = - \frac{x^2}{2} + C$$ - Multiply both sides by 2: $$y^2 + x^2 = 2C$$ - This is the equation of a family of circles centered at origin with radius $\sqrt{2C}$. 10. **Show the differential equation is homogeneous and solve:** - Given: $$\frac{dy}{dx} = -\frac{x^2 - 3y^2}{2xy}$$ - Check homogeneity: - Numerator and denominator are homogeneous functions of degree 2. - Substitute $v = \frac{y}{x}$, so $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$. - Substitute into equation: $$v + x \frac{dv}{dx} = -\frac{1 - 3v^2}{2v}$$ - Rearrange: $$x \frac{dv}{dx} = -\frac{1 - 3v^2}{2v} - v = -\frac{1 - 3v^2 + 2v^2}{2v} = -\frac{1 - v^2}{2v}$$ - Separate variables: $$\frac{2v}{v^2 - 1} dv = \frac{dx}{x}$$ - Integrate: $$\int \frac{2v}{v^2 - 1} dv = \int \frac{dx}{x}$$ - Use substitution $u = v^2 - 1$, $du = 2v dv$: $$\int \frac{du}{u} = \ln|x| + C$$ - So: $$\ln|v^2 - 1| = \ln|x| + C$$ - Exponentiate: $$|v^2 - 1| = K|x|$$ - Substitute back $v = \frac{y}{x}$: $$\left(\frac{y}{x}\right)^2 - 1 = Kx$$ - Multiply both sides by $x^2$: $$y^2 - x^2 = Kx^3$$ - This is the implicit solution. **Final answers:** - (a) Order = highest derivative order, Degree = power of highest derivative, Linear means derivatives and $y$ appear to power 1. - (b) Nonlinear, order 3, degree 32. - (c) Exact because $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. - (d) $y = C_1 e^{2.5x} + C_2 e^{0.5x}$. - (e) $\mathcal{L}\{1\} = \frac{1}{s}$. - (f) Family of circles: $x^2 + y^2 = 2C$. - (g) Homogeneous and solution: $y^2 - x^2 = Kx^3$.