Let’s provide detailed, step-by-step solutions for each question. --- ### **QUESTION 1** #### Equation: \[ \frac{dy}{dx} - y = 2(x) \] #### **Type:** This is a **first-order linear differential equation** of the form: \[ \frac{dy}{dx} + P(x)y = Q(x) \] where \(P(x) = -1\) and \(Q(x) = 2(x)\). --- ### **QUESTION 2** #### Equation: \[ 2x\frac{dy}{dx} - 2y = x \tan \left(\frac{y}{x}\right) \] with substitution \(y = vx\). #### **Type:** This is a **homogeneous equation** (can be written as a function of \(\frac{y}{x}\)). --- ### **QUESTION 3** #### Equation: \[ 2x\frac{dy}{dx} - 2y \tan z = y^2 \tan^2 x \] #### **Type:** This is a **nonlinear first-order equation** due to the \(y^2\) and \(\tan^2 x\) terms. --- ### **QUESTION 4** #### Equation: \[ x\frac{dz}{dx} + y = y^2x^2e^{2x} \] #### **Type:** This is a **first-order linear equation** in \(z\). --- ### **QUESTION 5** #### Equation: \[ (y-x^2)\frac{dy}{dx} - 2xy = 0 \] #### **Type:** This is a **homogeneous equation** (can be written as a function of \(\frac{y}{x}\)). --- ## **QUESTION 6** ### Solve the First Order Differential Equation \[ \frac{dy}{dx} = \frac{y^2}{3x^2 - x^3} \] #### **(a) As a homogeneous equation** This is a **separable differential equation**. **Step 1:** Separate the variables: \[ \frac{dy}{y^2} = \frac{dx}{3x^2 - x^3} \] **Step 2:** Integrate both sides: - Left: \(\int y^{-2} dy = -y^{-1}\) - Right: Factor denominator: \[ 3x^2 - x^3 = x^2(3 - x) \] So, \[ \int \frac{dx}{x^2(3 - x)} \] Use partial fractions: \[ \frac{1}{x^2(3-x)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{3-x} \] Multiply both sides by \(x^2(3-x)\): \[ 1 = A x (3-x) + B (3-x) + C x^2 \] Expand and collect terms: \[ 1 = A(3x - x^2) + B(3 - x) + C x^2 \] \[ 1 = (3A + 3B)x + (-A + C)x^2 + (3B) \] Let's expand carefully: Multiply out: \[ 1 = A x (3 - x) + B (3 - x) + C x^2 \] \[ = 3A x - A x^2 + 3B - B x + C x^2 \] \[ = (3B) + (3A x - B x) + (-A x^2 + C x^2) \] \[ = 3B + (3A - B)x + (C-A)x^2 \] Set equal to 1: Therefore, \[ 3B = 1 \implies B = \frac{1}{3} \] \[ 3A - B = 0 \implies 3A = B \implies A = \frac{B}{3} = \frac{1}{9} \] \[ C - A = 0 \implies C = A = \frac{1}{9} \] So, the decomposition is: \[ \frac{1}{x^2(3-x)} = \frac{1}{9}\frac{1}{x} + \frac{1}{3}\frac{1}{x^2} + \frac{1}{9}\frac{1}{3-x} \] **Step 3:** Integrate: \[ \int \frac{1}{x^2(3-x)} dx = \frac{1}{9} \int \frac{1}{x} dx + \frac{1}{3} \int \frac{1}{x^2} dx + \frac{1}{9} \int \frac{1}{3-x} dx \] \[ = \frac{1}{9} \ln |x| - \frac{1}{3x} - \frac{1}{9} \ln |3-x| + C \] **Step 4:** Write full solution: \[ -y^{-1} = \frac{1}{9} \ln |x| - \frac{1}{3x} - \frac{1}{9} \ln |3-x| + C \] Or, \[ \boxed{ \frac{-1}{y} = \frac{1}{9} \ln|x| - \frac{1}{3x} - \frac{1}{9} \ln|3-x| + C } \] --- #### **(b) As an exact equation** First, rewrite: \[ \frac{dy}{dx} - \frac{y^2}{3x^2 - x^3} = 0 \] Or, \[ (3x^2 - x^3)dy - y^2 dx = 0 \] Let \(M(x, y) = -y^2\), \(N(x, y) = 3x^2 - x^3\). Check for exactness: \[ \frac{\partial M}{\partial x} = 0 \] \[ \frac{\partial N}{\partial y} = 0 \] So, it **is exact**. Form the potential function \(F(x, y)\): \[ \frac{\partial F}{\partial x} = -y^2 \implies F(x, y) = -y^2x + h(y) \] \[ \frac{\partial F}{\partial y} = -2xy + h'(y) \] Set equal to \(N(x, y) = 3x^2 - x^3\): But this does not match, so equation as written is **not exact**. You would need an integrating factor, but the separable solution above is the simplest approach. --- ## **Summary Table** | Question | Type of Equation | Solution Steps Explained | |----------|----------------------------|----------------------------------------------------------| | 1 | First-order linear | Standard integrating factor method | | 2 | Homogeneous | Use substitution \(y = vx\) | | 3 | Nonlinear first-order | Nonlinear (due to \(y^2\)) | | 4 | First-order linear | Standard integrating factor method | | 5 | Homogeneous | Use substitution \(y = vx\) | | 6 | Separable | Separable, solved above | If you want solutions carried out for any other questions, let me know!

Certainly! Here's a detailed, step-by-step solution for each question, including calculations and the final answers. --- ## **QUESTION 1** **Equation:** \[ \frac{dy}{dx} - y = 2x \] ### **Step 1:** Recognize the type This is a **first-order linear differential equation** of the form: \[ \frac{dy}{dx} + P(x) y = Q(x) \] where \( P(x) = -1 \) and \( Q(x) = 2x \). ### **Step 2:** Find the integrating factor (IF) \[ \text{IF} = e^{\int P(x) dx} = e^{\int -1 dx} = e^{-x} \] ### **Step 3:** Multiply through by the integrating factor \[ e^{-x} \frac{dy}{dx} - e^{-x} y = 2x e^{-x} \] which simplifies to: \[ \frac{d}{dx} [ y e^{-x} ] = 2x e^{-x} \] ### **Step 4:** Integrate both sides \[ \int \frac{d}{dx} [ y e^{-x} ] dx = \int 2x e^{-x} dx \] \[ y e^{-x} = \int 2x e^{-x} dx + C \] ### **Step 5:** Compute \(\int 2x e^{-x} dx\) Use **integration by parts**: - Let \( u = 2x \Rightarrow du = 2 dx \) - Let \( dv = e^{-x} dx \Rightarrow v = - e^{-x} \) Then: \[ \int 2x e^{-x} dx = uv - \int v du = -2x e^{-x} - \int -2 e^{-x} dx \] \[ = -2x e^{-x} + 2 \int e^{-x} dx = -2x e^{-x} + 2(- e^{-x}) + C \] \[ = -2x e^{-x} - 2 e^{-x} + C \] ### **Step 6:** Write the solution \[ y e^{-x} = -2x e^{-x} - 2 e^{-x} + C \] Multiply both sides by \( e^{x} \): \[ y = -2x - 2 + C e^{x} \] ### **Final answer:** \[ \boxed{ \boxed{ y = C e^{x} - 2x - 2 } } \] where \( C \) is an arbitrary constant. --- ## **QUESTION 2** **Equation:** \[ 2x \frac{dy}{dx} - 2 y = x \tan \left( \frac{y}{x} \right) \] with substitution \( y = v x \). ### **Step 1:** Substitution \( y = v x \) Then: \[ dy/dx = v + x dv/dx \] ### **Step 2:** Rewrite the equation Substitute into the original: \[ 2x (v + x dv/dx) - 2 v x = x \tan(v) \] Simplify: \[ 2x v + 2 x^2 dv/dx - 2 v x = x \tan v \] Note that \( 2x v - 2 v x = 0 \), so these cancel out: \[ 2 x^2 dv/dx = x \tan v \] Divide both sides by \( x \): \[ 2 x dv/dx = \tan v \] or \[ dv/dx = \frac{\tan v}{2x} \] ### **Step 3:** Separate variables \[ \frac{dv}{\tan v} = \frac{1}{2x} dx \] Recall: \[ \frac{1}{\tan v} = \cot v \] So: \[ \cot v dv = \frac{1}{2x} dx \] ### **Step 4:** Integrate both sides \[ \int \cot v dv = \int \frac{1}{2x} dx \] - \(\int \cot v dv = \ln |\sin v| + C_1\) - \(\int \frac{1}{2x} dx = \frac{1}{2} \ln |x| + C_2\) Combine constants: \[ \ln |\sin v| = \frac{1}{2} \ln |x| + C \] ### **Step 5:** Write solution in terms of \( y \) Recall \( v = y/x \): \[ \sin v = K x^{1/2} \] where \( K = e^{C} \). Substituting back: \[ \sin \left( \frac{y}{x} \right) = K x^{1/2} \] ### **Final answer:** \[ \boxed{ \sin \left( \frac{y}{x} \right) = C x^{1/2} } \] where \( C \) is an arbitrary constant. --- ## **QUESTION 3** **Equation:** \[ 2x \frac{dy}{dx} - 2 y \tan z = y^2 \tan^2 x \] *Assuming the problem is to solve for \( y \) with respect to \( x \).* --- *(Note: The original question appears incomplete or contains a typo; assuming it is similar in form to previous equations, and that \( z \) is a parameter or function. If \( z \) is constant, the equation reduces to a Bernoulli or nonlinear form. Without further clarification, let's skip detailed solution here.)* --- ## **QUESTION 4** **Equation:** \[ x \frac{dz}{dx} + y = y^2 x^2 e^{2x} \] ### **Step 1:** Isolate \( dz/dx \) \[ x \frac{dz}{dx} = y^2 x^2 e^{2x} - y \] \[ \frac{dz}{dx} = \frac{ y^2 x^2 e^{2x} - y }{x} \] ### **Step 2:** Recognize as a linear equation in \( z \) *But since \( y \) is involved, unless \( y \) is known or expressed in terms of \( x \), this is a coupled problem.* --- ## **QUESTION 5** **Equation:** \[ (y - x^2) \frac{dy}{dx} - 2xy = 0 \] ### **Step 1:** Recognize as homogeneous Rewrite: \[ (y - x^2) \frac{dy}{dx} = 2xy \] Divide both sides by \( y - x^2 \): \[ \frac{dy}{dx} = \frac{2xy}{ y - x^2 } \] ### **Step 2:** Use substitution \( y = v x \) \[ dy/dx = v + x dv/dx \] Substitute: \[ v + x dv/dx = \frac{2x v x}{v x - x^2} = \frac{2 v x^2}{x (v - x)} = \frac{2 v x}{v - x} \] Rearranged: \[ v + x dv/dx = \frac{2 v x}{v - x} \] Solve for \( dv/dx \): \[ x dv/dx = \frac{2 v x}{v - x} - v \] Express numerator: \[ x dv/dx = \frac{2 v x - v (v - x)}{v - x} \] Simplify numerator: \[ 2 v x - v^2 + v x = 3 v x - v^2 \] So: \[ x dv/dx = \frac{3 v x - v^2}{v - x} \] --- ## **QUESTION 6** **Equation:** \[ \frac{dy}{dx} = \frac{y^2}{3x^2 - x^3} \] ### **Part (a):** As a **separable differential equation** **Step 1:** Separate variables: \[ \frac{dy}{y^2} = \frac{dx}{3x^2 - x^3} \] **Step 2:** Integrate left side: \[ \int y^{-2} dy = - y^{-1} + C \] **Step 3:** Simplify the right side denominator: \[ 3x^2 - x^3 = x^2 (3 - x) \] so: \[ \int \frac{dx}{x^2 (3 - x)} \] **Step 4:** Partial fraction decomposition: \[ \frac{1}{x^2 (3 - x)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{3 - x} \] Multiply both sides by \( x^2 (3 - x) \): \[ 1 = A x (3 - x) + B (3 - x) + C x^2 \] Expand: \[ 1 = 3A x - A x^2 + 3 B - B x + C x^2 \] Group like terms: \[ 1 = (3 B) + (3A x - B x) + (-A x^2 + C x^2) \] \[ 1 = 3B + (3A - B) x + (C - A) x^2 \] Set coefficients equal: - Constant term: \[ 3 B = 1 \Rightarrow B = \frac{1}{3} \] - Coefficient of \( x \): \[ 3A - B = 0 \Rightarrow 3A = B = \frac{1}{3} \Rightarrow A = \frac{1}{9} \] - Coefficient of \( x^2 \): \[ C - A = 0 \Rightarrow C = A = \frac{1}{9} \] ### **Step 5:** Integrate right side \[ \int \left( \frac{1/9}{x} + \frac{1/3}{x^2} + \frac{1/9}{3 - x} \right) dx \] which gives: \[ \frac{1}{9} \ln|x| - \frac{1}{3 x} + \frac{1}{9} \ln|3 - x| + C \] ### **Step 6:** Write the solution: \[ - \frac{1}{y} = \frac{1}{9} \ln|x| - \frac{1}{3 x} + \frac{1}{9} \ln|3 - x| + C \] or equivalently: \[ \boxed{ \frac{1}{y} = - \left( \frac{1}{9} \ln|x| - \frac{1}{3 x} + \frac{1}{9} \ln|3 - x| \right) + C' } \] --- ## **Summary of Final Answers:** | Question | Final Solution | |------------|--------------------------------------------------------------| | 1 | \( y = C e^{x} - 2x - 2 \) | | 2 | \( \sin \left( \frac{y}{x} \right) = C x^{1/2} \) | | 6 (a) | \( \frac{1}{y} = - \left( \frac{1}{9} \ln|x| - \frac{1}{3 x} + \frac{1}{9} \ln|3 - x| \right) + C \) | --- If you need further clarification or solutions for the remaining parts, please let me know!

Certainly! Here are detailed calculations and solutions for **Questions 3, 4, and 5** based on the given differential equations. --- # **QUESTION 3** **Equation:** \[ 2x \frac{dy}{dx} - 2 y \tan z = y^2 \tan^2 x \] --- ### **Assumption / Clarification:** The question appears to involve a relation with \(z\). Typically, if \(z\) is a constant or a parameter, we can treat it as constant. Alternatively, if \(z = y\), then the equation would be different. **For this solution, assuming \(z\) is a constant**, the equation reduces to an ordinary differential equation (ODE) in \(y\). --- ### **Step 1:** Rewrite the equation Divide both sides by \(2x\): \[ \frac{dy}{dx} - \frac{y \tan z}{x} = \frac{y^2 \tan^2 x}{2x} \] This is a **Bernoulli differential equation** of the form: \[ \frac{dy}{dx} + P(x) y = Q(x) y^n \] where: - \( P(x) = - \frac{\tan z}{x} \) - \( Q(x) = \frac{\tan^2 x}{2x} \) - \( n = 2 \) --- ### **Step 2:** Standard Bernoulli substitution Let \( u = y^{1 - n} = y^{-1} \). Then: \[ \frac{dy}{dx} = - y^2 \frac{du}{dx} \] Substitute into the original: \[ - y^2 \frac{du}{dx} - \frac{y \tan z}{x} = \frac{y^2 \tan^2 x}{2x} \] Divide through by \( y^2 \): \[ - \frac{du}{dx} - \frac{\tan z}{x y} = \frac{\tan^2 x}{2x} \] But since \( u = 1/y \), then \( y = 1/u \), so: \[ - \frac{du}{dx} - \frac{\tan z}{x} u = \frac{\tan^2 x}{2x} \] Rearranged: \[ \frac{du}{dx} + \frac{\tan z}{x} u = - \frac{\tan^2 x}{2x} \] --- ### **Step 3:** Solve the linear first-order ODE in \( u \) Standard form: \[ \frac{du}{dx} + P(x) u = Q(x) \] with: - \( P(x) = \frac{\tan z}{x} \) - \( Q(x) = - \frac{\tan^2 x}{2x} \) ### **Step 4:** Find integrating factor: \[ \mu(x) = e^{\int P(x) dx} = e^{\int \frac{\tan z}{x} dx} = e^{\tan z \int \frac{1}{x} dx} = e^{\tan z \ln |x|} = |x|^{\tan z} \] ### **Step 5:** Write solution: \[ u(x) \times \mu(x) = \int Q(x) \mu(x) dx + C \] \[ u(x) |x|^{\tan z} = \int - \frac{\tan^2 x}{2x} |x|^{\tan z} dx + C \] --- ### **Step 6:** Final solution for \( y \): Recall \( u = 1/y \): \[ y = \frac{1}{u} = \frac{|x|^{\tan z}}{ \int - \frac{\tan^2 x}{2x} |x|^{\tan z} dx + C } \] --- ### **Note:** - The integral in the numerator involves \( \tan^2 x \). It may not have a simple closed form. Therefore, the solution is expressed in integral form unless further assumptions are made. --- # **SUMMARY for Question 3:** \[ \boxed{ \text{Solution: } y(x) = \frac{|x|^{\tan z}}{ \displaystyle \int - \frac{\tan^2 x}{2x} |x|^{\tan z} dx + C } } \] Where the integral may be evaluated numerically or left in integral form. --- # **QUESTION 4** **Equation:** \[ x \frac{dz}{dx} + y = y^2 x^2 e^{2x} \] --- ### **Step 1:** Express \( dz/dx \): \[ \frac{dz}{dx} = \frac{ y^2 x^2 e^{2x} - y }{ x } \] --- ### **Step 2:** Recognize whether this is a linear or nonlinear ODE: - \(z\) appears only through its derivative; \( y \) is involved explicitly. - If \( y \) is known or can be expressed in terms of \( x \), then \( z \) can be integrated accordingly. **Assumption / Simplification:** - Suppose \( y \) is a known function of \( x \) or constant. For demonstration, assume \( y \) is a constant \( y_0 \). --- ### **Step 3:** Solution for \(z\) when \( y \) is constant \[ \frac{dz}{dx} = \frac{ y_0^2 x^2 e^{2x} - y_0 }{ x } \] Integrate both sides: \[ z(x) = \int \frac{ y_0^2 x^2 e^{2x} - y_0 }{ x } dx + C \] Split integral: \[ z(x) = y_0^2 \int x e^{2x} dx - y_0 \int \frac{1}{x} dx + C \] --- ### **Step 4:** Compute integrals - First integral: \( \int x e^{2x} dx \) Use **integration by parts**: - \( u = x \Rightarrow du = dx \) - \( dv = e^{2x} dx \Rightarrow v = \frac{1}{2} e^{2x} \) \[ \int x e^{2x} dx = x \times \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} dx = \frac{x}{2} e^{2x} - \frac{1}{4} e^{2x} + C \] - Second integral: \[ \int \frac{1}{x} dx = \ln |x| + C \] ### **Step 5:** Final expression for \(z(x)\): \[ z(x) = y_0^2 \left( \frac{x}{2} e^{2x} - \frac{1}{4} e^{2x} \right) - y_0 \ln |x| + C \] --- ### **Summary:** \[ \boxed{ z(x) = y_0^2 \left( \frac{x}{2} e^{2x} - \frac{1}{4} e^{2x} \right) - y_0 \ln |x| + C } \] This solution holds under the assumption that \( y \) is a constant \( y_0 \). If \( y \) varies, the problem becomes more complex and requires additional information. --- # **QUESTION 5** **Equation:** \[ (y - x^2) \frac{dy}{dx} - 2 x y = 0 \] --- ### **Step 1:** Recognize as homogeneous Rewrite: \[ (y - x^2) \frac{dy}{dx} = 2 x y \] --- ### **Step 2:** Substitution \( y = v x \) \[ dy/dx = v + x dv/dx \] Replace in the equation: \[ (v x - x^2) (v + x dv/dx) = 2 x v x \] Simplify: \[ x(v - x) (v + x dv/dx) = 2 v x^2 \] Divide both sides by \( x \): \[ (v - x)(v + x dv/dx) = 2 v x \] --- ### **Step 3:** Expand left side \[ (v - x) v + (v - x) x \frac{dv}{dx} = 2 v x \] \[ v^2 - x v + x (v - x) \frac{dv}{dx} = 2 v x \] Bring all to one side: \[ x (v - x) \frac{dv}{dx} = 2 v x - v^2 + x v \] Note that: \[ 2 v x + x v = 3 v x \] So, \[ x (v - x) \frac{dv}{dx} = 3 v x - v^2 \] --- ### **Step 4:** Divide through by \( x (v - x) \) \[ \frac{dv}{dx} = \frac{3 v x - v^2}{x (v - x)} \] which simplifies to: \[ \frac{dv}{dx} = \frac{v(3 x - v)}{x (v - x)} \] --- ### **Optional:** Simplify further or proceed with substitution. --- # **Summary:** The differential equation reduces to: \[ \boxed{ \frac{dv}{dx} = \frac{v(3 x - v)}{x (v - x)} } \] which can be further solved via substitution or partial fractions, depending on the method preferred. --- # **Final notes:** - For all equations, assumptions about parameters (like \(z\) or \(y\)) being constants or functions influence the solution method. - Some equations lead to Bernoulli or linear first-order equations, solvable with standard methods. - Integrals involving \( \tan x \) or exponential functions require integration by parts. --- If you'd like me to proceed with complete solutions for any of these equations or clarify specific parts, please let me know!

Certainly! Let's carefully revisit **Question 6 (part b)** and derive its explicit solution. --- ### **Question 6 (Part b):** Given differential equation: \[ \frac{dy}{dx} = \frac{y^2}{3x^2 - x^3} \] --- ### **Step 1: Recognize the equation as separable** Rewrite: \[ \frac{dy}{dx} = \frac{y^2}{x^2 (3 - x)} \] Separate variables: \[ \frac{dy}{y^2} = \frac{dx}{x^2 (3 - x)} \] --- ### **Step 2: Integrate both sides** - Left side: \[ \int y^{-2} dy = - y^{-1} + C \] - Right side: \[ \int \frac{dx}{x^2 (3 - x)} \] --- ### **Step 3: Partial fractions of the right side** Express: \[ \frac{1}{x^2 (3 - x)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{3 - x} \] Multiply both sides by \( x^2 (3 - x) \): \[ 1 = A x (3 - x) + B (3 - x) + C x^2 \] Expand: \[ 1 = 3A x - A x^2 + 3 B - B x + C x^2 \] Group by powers of \( x \): \[ 1 = (3 B) + (3 A x - B x) + (-A x^2 + C x^2) \] Matching coefficients: - Constant term: \[ 3 B = 1 \Rightarrow B = \frac{1}{3} \] - \( x \) term: \[ 3 A - B = 0 \Rightarrow 3A = B = \frac{1}{3} \Rightarrow A = \frac{1}{9} \] - \( x^2 \) term: \[ C - A = 0 \Rightarrow C = A = \frac{1}{9} \] --- ### **Step 4: Write the integral** So, \[ \int \frac{dx}{x^2 (3 - x)} = \int \left( \frac{1/9}{x} + \frac{1/3}{x^2} + \frac{1/9}{3 - x} \right) dx \] which simplifies to: \[ \frac{1}{9} \int \frac{1}{x} dx + \frac{1}{3} \int \frac{1}{x^2} dx + \frac{1}{9} \int \frac{1}{3 - x} dx \] --- ### **Step 5: Compute each integral** - \(\int \frac{1}{x} dx = \ln |x|\) - \(\int \frac{1}{x^2} dx = - \frac{1}{x}\) - \(\int \frac{1}{3 - x} dx = - \ln |3 - x|\) Thus, \[ \int \frac{dx}{x^2 (3 - x)} = \frac{1}{9} \ln |x| - \frac{1}{3x} - \frac{1}{9} \ln |3 - x| + C \] --- ### **Step 6: Write the general solution** Recall: \[ - y^{-1} = \frac{1}{9} \ln |x| - \frac{1}{3 x} - \frac{1}{9} \ln |3 - x| + C \] Multiply both sides by \(-1\): \[ \frac{1}{y} = - \left( \frac{1}{9} \ln |x| - \frac{1}{3 x} - \frac{1}{9} \ln |3 - x| \right) + C' \] or equivalently, \[ \boxed{ \frac{1}{y} = - \frac{1}{9} \ln |x| + \frac{1}{3 x} + \frac{1}{9} \ln |3 - x| + C' } \] where \( C' \) is an arbitrary constant. --- ### **Final explicit solution:** \[ \boxed{ \boxed{ y(x) = \frac{1}{ - \frac{1}{9} \ln |x| + \frac{1}{3 x} + \frac{1}{9} \ln |3 - x| + C } } \] --- ### **Summary:** - The solution involves logarithmic functions and rational terms. - The constant \( C \) accounts for the integration constant. --- If you need further clarification or specific evaluations, let me know!

Certainly! Here are the final answers for each question: ### **QUESTION 1** \[ \boxed{ y = C e^{x} - 2x - 2 } \] ### **QUESTION 2** \[ \boxed{ \sin \left( \frac{y}{x} \right) = C x^{1/2} } \] ### **QUESTION 3** \[ \boxed{ y(x) = \frac{|x|^{\tan z}}{ \displaystyle \int - \frac{\tan^2 x}{2x} |x|^{\tan z} dx + C } } \] *(Integral may be left unevaluated or computed numerically)* ### **QUESTION 4** \[ \boxed{ z(x) = y_0^2 \left( \frac{x}{2} e^{2x} - \frac{1}{4} e^{2x} \right) - y_0 \ln |x| + C } \] *(Assuming \( y = y_0 \) constant)* ### **QUESTION 5** \[ \boxed{ \text{Solution involves substitution } y = v x \text{ leading to a differential equation in } v } \] *(Further solving requires additional steps; key form: \(\frac{dv}{dx} = \frac{v(3 x - v)}{x (v - x)}\))* ### **QUESTION 6 (Part b)** \[ \boxed{ y(x) = \frac{1}{ - \frac{1}{9} \ln |x| + \frac{1}{3 x} + \frac{1}{9} \ln |3 - x| + C } } \] --- Let me know if you'd like the detailed steps for any of these!