Let's solve each differential equation step by step: --- ### 1. \( y' + 2y = x^3 \) **Step 1: Identify the type.** This is a first-order linear ordinary differential equation. **Step 2: Write in standard form.** \( y' + 2y = x^3 \) **Step 3: Find the integrating factor (IF).** \( IF = e^{\int 2 dx} = e^{2x} \) **Step 4: Multiply both sides by IF.** \( e^{2x}y' + 2e^{2x}y = x^3 e^{2x} \) **Step 5: Left side becomes derivative:** \( \frac{d}{dx}(e^{2x}y) = x^3 e^{2x} \) **Step 6: Integrate both sides:** \( e^{2x}y = \int x^3 e^{2x} dx \) Use integration by parts (or formula): Let \( u = x^3, \ dv = e^{2x}dx \) Repeated integration by parts gives: \[ \int x^3 e^{2x} dx = \frac{e^{2x}}{2} x^3 - \frac{3e^{2x}}{4} x^2 + \frac{3e^{2x}}{4} x - \frac{3e^{2x}}{8} + C \] **Step 7: Solve for \(y\):** \[ y = \frac{1}{e^{2x}} \left( \frac{e^{2x}}{2} x^3 - \frac{3e^{2x}}{4} x^2 + \frac{3e^{2x}}{4} x - \frac{3e^{2x}}{8} + C \right) \] \[ \boxed{y = \frac{1}{2}x^3 - \frac{3}{4}x^2 + \frac{3}{4}x - \frac{3}{8} + Ce^{-2x}} \] --- ### 2. \( y' - \tan(x)y = x \) **Step 1: Standard form:** \( y' - \tan(x)y = x \) **Step 2: Integrating factor:** \( IF = e^{\int -\tan(x) dx} = e^{-\ln|\sec x|} = \cos x \) **Step 3: Multiply by IF:** \( \cos x y' - \sin x y = x \cos x \) Left side: \( \frac{d}{dx}(\cos x\, y) = x\cos x \) **Step 4: Integrate both sides:** \( \cos x\, y = \int x\cos x dx \) Integrate by parts: Let \( u = x, dv = \cos x dx \to du = dx, v = \sin x \) \[ \int x\cos x dx = x\sin x - \int \sin x dx = x\sin x + \cos x \] **Step 5: Solve for y:** \[ \cos x\, y = x\sin x + \cos x + C \] \[ \boxed{y = x\tan x + 1 + C\sec x} \] --- ### 3. \( y' + \frac{1}{x}y = y^2 \) This is **Bernoulli's equation**: \[ y' + P(x)y = Q(x)y^n \] Here, \( n=2 \). Let \( v = y^{1-n} = y^{-1} \), \( y = v^{-1} \), \( y' = -v^{-2} v' \): Plug into original: \[ -y^{-2} v' + \frac{1}{x} y = y^2 \] \[ - v' + \frac{1}{x} v = 1 \] \[ v' - \frac{1}{x} v = -1 \] Now, linear in \( v \): **IF:** \( e^{-\int \frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x} \) Multiply both sides: \[ \frac{1}{x} v' - \frac{1}{x^2} v = -\frac{1}{x} \] \[ \frac{d}{dx} \left( \frac{v}{x} \right) = -\frac{1}{x} \] Integrate both sides: \[ \frac{v}{x} = -\ln|x| + C \] \[ v = -x\ln|x| + Cx \] Recall \( v = 1/y \): \[ \boxed{y = \frac{1}{-x\ln|x| + Cx}} \] --- ### 4. \( (2x + y)dx + (x - y^2)dy = 0 \) **Step 1:** Write as \( M(x,y)dx + N(x,y)dy = 0 \), where \( M = 2x+y \), \( N = x-y^2 \) Check exactness: \[ \frac{\partial M}{\partial y} = 1,\quad \frac{\partial N}{\partial x} = 1 \] So, **exact**. Find \( F(x, y) \) such that \( F_x = 2x + y, F_y = x - y^2 \): Integrate \( F_x \) wrt \( x \): \[ F(x, y) = \int (2x + y)dx = x^2 + xy + h(y) \] Differentiate wrt \( y \): \[ F_y = x + h'(y) \] Set equal to \( N(x, y) \): \[ x + h'(y) = x - y^2 \implies h'(y) = -y^2 \implies h(y) = -\frac{y^3}{3} \] So, \[ F(x, y) = x^2 + xy - \frac{y^3}{3} = C \] \[ \boxed{x^2 + xy - \frac{y^3}{3} = C} \] --- ### 5. \( (2x + xy)dx + xdy = 0 \) \( M(x, y) = 2x + xy \), \( N(x, y) = x \) Check exactness: \[ M_y = x,\quad N_x = 1 \] Not exact. Check integrating factor (try function of \( y \)): \[ \frac{N_x - M_y}{M} = \frac{1 - x}{2x + xy} \] Not a function of \( y \) alone. Try function of \( x \): \[ \frac{M_y - N_x}{N} = \frac{x - 1}{x} = 1 - \frac{1}{x} \] Not a function of \( x \) alone. Let’s try to make it separable: Rewrite: \[ (2x + xy)dx + xdy = 0 \implies (2x + xy)dx = -xdy \] \[ (2x + xy)dx + xdy = 0 \] \[ (2x + xy)dx = -xdy \] \[ \frac{2x + xy}{x}dx = -dy \] \[ (2 + y)dx = -dy \] \[ (2 + y)dx + dy = 0 \] Integrate both sides: \[ \int (2 + y)dx + \int dy = 0 \] But \( y \) is a function of \( x \), so separate variables: From \( (2 + y)dx = -dy \): \[ \frac{dy}{2 + y} = -dx \] Integrate: \[ \int \frac{1}{2 + y} dy = -\int dx \] \[ \ln|2 + y| = -x + C \] \[ |2 + y| = e^{-x + C} = Ce^{-x} \] \[ \boxed{y = C e^{-x} - 2} \] --- ## **Final Answers** 1. **\( y' + 2y = x^3 \):** \[ \boxed{y = \frac{1}{2}x^3 - \frac{3}{4}x^2 + \frac{3}{4}x - \frac{3}{8} + Ce^{-2x}} \] 2. **\( y' - \tan(x)y = x \):** \[ \boxed{y = x\tan x + 1 + C\sec x} \] 3. **\( y' + \frac{1}{x}y = y^2 \):** \[ \boxed{y = \frac{1}{-x\ln|x| + Cx}} \] 4. **\( (2x + y)dx + (x - y^2)dy = 0 \):** \[ \boxed{x^2 + xy - \frac{y^3}{3} = C} \] 5. **\( (2x + xy)dx + xdy = 0 \):** \[ \boxed{y = Ce^{-x} - 2} \] If you need detailed integration steps for any particular equation or want to see the solution process for any one, let me know!