Let's solve each differential equation step by step: --- ### 1. \(\frac{dy}{dx} = \frac{y}{1 + x^2}\) **Step 1: Separate variables** \[ \frac{dy}{y} = \frac{dx}{1 + x^2} \] **Step 2: Integrate both sides** \[ \int \frac{1}{y} dy = \int \frac{1}{1 + x^2} dx \] \[ \ln|y| = \arctan(x) + C \] **Step 3: Solve for \(y\)** \[ |y| = e^{\arctan(x) + C} = Ae^{\arctan(x)} \] **Final Answer:** \[ \boxed{y = Ae^{\arctan(x)}} \] --- ### 2. \(\frac{dy}{dx} = \frac{y}{1 - x^2}\) **Step 1: Separate variables** \[ \frac{dy}{y} = \frac{dx}{1-x^2} \] **Step 2: Integrate both sides** \[ \int \frac{1}{y} dy = \int \frac{1}{1-x^2} dx \] \[ \ln|y| = \frac{1}{2} \ln\left|\frac{1+x}{1-x}\right| + C \] **Step 3: Solve for \(y\)** \[ |y| = A\sqrt{\frac{1+x}{1-x}} \] **Final Answer:** \[ \boxed{y = A\sqrt{\frac{1+x}{1-x}}} \] --- ### 3. \(\frac{dy}{dx} = yx^2 e^x\) **Step 1: Separate variables** \[ \frac{dy}{y} = x^2 e^x dx \] **Step 2: Integrate both sides** \[ \int \frac{1}{y} dy = \int x^2 e^x dx \] Let’s solve the right side using integration by parts: Let \(u = x^2\), \(dv = e^x dx\), then \(du = 2x dx\), \(v = e^x\). \[ \int x^2 e^x dx = x^2 e^x - \int 2x e^x dx \] Now, \(\int 2x e^x dx = 2 (x e^x - \int e^x dx) = 2(x e^x - e^x)\) So, \[ \int x^2 e^x dx = x^2 e^x - 2x e^x + 2 e^x \] So, \[ \ln |y| = x^2 e^x - 2x e^x + 2e^x + C \] **Step 3: Solve for \(y\)** \[ y = Ae^{x^2 e^x - 2x e^x + 2e^x} \] **Final Answer:** \[ \boxed{y = Ae^{x^2 e^x - 2x e^x + 2e^x}} \] --- ### 4. \(\frac{dy}{dx} = \frac{x^2 + xy}{x^2 - y}\) This is a **homogeneous** equation. Let’s use the substitution \(y = vx\), \(dy = v dx + x dv\). \[ \frac{dy}{dx} = \frac{x^2 + x(vx)}{x^2 - vx} = \frac{x^2 + vx^2}{x^2 - vx} = \frac{x^2(1 + v)}{x^2(1 - v)} = \frac{1 + v}{1 - v} \] But \(\frac{dy}{dx} = v + x \frac{dv}{dx}\): \[ v + x \frac{dv}{dx} = \frac{1 + v}{1 - v} \] \[ x \frac{dv}{dx} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v(1 - v)}{1 - v} = \frac{1 + v - v + v^2}{1-v} = \frac{1 + v^2}{1-v} \] \[ \frac{1-v}{1+v^2} dv = \frac{dx}{x} \] Break up the left side: \[ \int \frac{1}{1+v^2} dv - \int \frac{v}{1+v^2} dv = \int \frac{dx}{x} \] The first term is \(\arctan v\), the second is \(\frac{1}{2} \ln(1+v^2)\): \[ \arctan v - \frac{1}{2} \ln(1+v^2) = \ln|x| + C \] Recall \(v = \frac{y}{x}\): \[ \arctan\left(\frac{y}{x}\right) - \frac{1}{2} \ln \left(1 + \left(\frac{y}{x}\right)^2\right) = \ln|x| + C \] **Final Answer:** \[ \boxed{\arctan\left(\frac{y}{x}\right) - \frac{1}{2} \ln \left(1 + \left(\frac{y}{x}\right)^2\right) = \ln|x| + C} \] --- (If you want the remaining problems solved, please let me know! Or specify any particular one.)

Certainly! Let's continue solving the remaining differential equations step-by-step. --- ### 5. \(\frac{dy}{dx} = 5 \tan y\) **Step 1: Separate variables** \[ \frac{dy}{\tan y} = 5 dx \] Recall that \(\frac{1}{\tan y} = \cot y\), so: \[ \cot y\, dy = 5 dx \] **Step 2: Integrate both sides** \[ \int \cot y\, dy = \int 5\, dx \] The integral of \(\cot y\) is \(\ln|\sin y|\). So: \[ \ln|\sin y| = 5x + C \] **Step 3: Solve for \(y\)** \[ |\sin y| = Ae^{5x} \] where \(A = e^{C}\). The general solution: \[ \boxed{\sin y = \pm A e^{5x}} \] --- ### 6. \( y'' + y = 0 \) This is a second-order linear homogeneous differential equation with constant coefficients. **Step 1: Characteristic equation** \[ r^2 + 1 = 0 \] \[ r^2 = -1 \] \[ r = \pm i \] **Step 2: General solution** \[ y(x) = C_1 \cos x + C_2 \sin x \] --- ### 7. \(\frac{dy}{dx} + x y = 0\) This is a first-order linear differential equation. **Step 1: Write in standard form** \[ \frac{dy}{dx} + x y = 0 \] **Step 2: Find integrating factor** \[ \mu(x) = e^{\int x dx} = e^{\frac{x^2}{2}} \] **Step 3: Multiply both sides by \(\mu(x)\)** \[ e^{\frac{x^2}{2}} \frac{dy}{dx} + x e^{\frac{x^2}{2}} y = 0 \] which simplifies to: \[ \frac{d}{dx}\left( y e^{\frac{x^2}{2}} \right) = 0 \] **Step 4: Integrate** \[ y e^{\frac{x^2}{2}} = C \] \[ \boxed{y = C e^{-\frac{x^2}{2}}} \] --- ### 8. \(\frac{dy}{dx} = y^n\), where \(n \neq 1\) **Note:** The problem seems incomplete, but assuming this is the general form, let's solve: \[ \frac{dy}{dx} = y^n \] **Step 1: Separate variables** \[ \frac{dy}{y^n} = dx \] which is: \[ y^{-n} dy = dx \] **Step 2: Integrate** \[ \int y^{-n} dy = \int dx \] \[ \frac{y^{1 - n}}{1 - n} = x + C \] **Final solution:** \[ \boxed{ y^{1 - n} = (1 - n)(x + C) } \] *(valid for \(n \neq 1\)).* --- **Summary of remaining solutions:** | Differential Equation | General Solution | |------------------------|-------------------| | \( y'' + y = 0 \) | \( y = C_1 \cos x + C_2 \sin x \) | | \( \frac{dy}{dx} + x y = 0 \) | \( y = C e^{-\frac{x^2}{2}} \) | | \( \frac{dy}{dx} = y^n \), \( n \neq 1 \) | \( y^{1 - n} = (1 - n)(x + C) \) | --- If you have any more questions or need further clarifications, feel free to ask!

Certainly! Here are the final answers: 1. \( y = Ae^{\arctan(x)} \) 2. \( y = A \sqrt{\frac{1+x}{1-x}} \) 3. \( y = Ae^{x^2 e^x - 2x e^x + 2 e^x} \) 4. \( \arctan\left(\frac{y}{x}\right) - \frac{1}{2} \ln \left(1 + \left(\frac{y}{x}\right)^2\right) = \ln|x| + C \) 5. \( \sin y = \pm A e^{5x} \) 6. \( y = C_1 \cos x + C_2 \sin x \) 7. \( y = C e^{-\frac{x^2}{2}} \) 8. \( y^{1 - n} = (1 - n)(x + C) \) (for \( n \neq 1 \)) Let me know if you need further assistance!