Let's solve each problem step-by-step! --- ## **1. By considering \( y = u x \), solve the equation:** \[ y^4 dx + 2(x^3 - x y^3) dy = 0 \] ### **Step 1: Substitute \( y = u x \)** Let \( y = u x \implies dy = u dx + x du \). Replace \( y \) and \( dy \) in the equation: - \( y^4 = (u x)^4 = u^4 x^4 \) - \( x^3 - x y^3 = x^3 - x (u x)^3 = x^3 - x u^3 x^3 = x^3 - u^3 x^4 \) So the equation becomes: \[ u^4 x^4 dx + 2[x^3 - u^3 x^4](u dx + x du) = 0 \] ### **Step 2: Expand** Expand the terms: \[ u^4 x^4 dx + 2[x^3 u dx + x^4 u du - u^3 x^4 u dx - u^3 x^4 x du] = 0 \] \[ u^4 x^4 dx + 2[x^3 u dx + x^4 u du - u^4 x^4 dx - u^3 x^5 du] = 0 \] Further simplifying: \[ u^4 x^4 dx + 2 x^3 u dx + 2 x^4 u du - 2 u^4 x^4 dx - 2 u^3 x^5 du = 0 \] Group like terms: \[ (u^4 x^4 - 2 u^4 x^4) dx + 2 x^3 u dx + (2 x^4 u du - 2 u^3 x^5 du) = 0 \] \[ (-u^4 x^4 + 2 x^3 u) dx + 2 x^4 u (1 - u^2 x) du = 0 \] ### **Step 3: Separate Variables** Bring all \( dx \) terms to one side and \( du \) to the other: \[ \frac{-u^4 x^4 + 2 x^3 u}{2 x^4 u (1-u^2 x)} dx + du = 0 \] \[ \frac{-u^4 x^4 + 2 x^3 u}{2 x^4 u (1-u^2 x)} dx = -du \] Let's factor numerator: \[ -u^4 x^4 + 2 x^3 u = x^3 u (-u^3 x + 2) \] So, \[ \frac{x^3 u (-u^3 x + 2)}{2 x^4 u (1-u^2 x)} dx = -du \] \[ \frac{-u^3 x + 2}{2 x (1-u^2 x)} dx = -du \] \[ \frac{-u^3 x + 2}{2 x (1-u^2 x)} dx + du = 0 \] ### **Step 4: Integrate** Let’s write as: \[ \frac{-u^3 x + 2}{2 x (1-u^2 x)} dx = -du \] This is a bit complex, but let's try to integrate both sides. Let’s set \( v = 1-u^2 x \implies dv = -u^2 dx \implies dx = \frac{dv}{-u^2} \), but the numerator is more complicated. Alternatively, let's try integrating as is: Let’s treat \( u \) as a constant with respect to \( x \) for the integration: Let’s try partial fractions: \[ \frac{-u^3 x + 2}{2 x (1-u^2 x)} \] Let’s write: \[ \frac{-u^3 x + 2}{2 x (1-u^2 x)} = \frac{A}{x} + \frac{B}{1-u^2 x} \] Multiply both sides by \( 2x(1-u^2 x) \): \[ -u^3 x + 2 = 2A(1-u^2 x) + 2B x \] Expand: \[ -u^3 x + 2 = 2A - 2A u^2 x + 2B x \] \[ -u^3 x + 2 = 2A + (2B - 2A u^2)x \] Match coefficients: - Constant: \(2A = 2 \implies A = 1\) - \(x\) term: \(-u^3 = 2B - 2A u^2\) - Substitute \(A=1\): \(-u^3 = 2B - 2u^2\) \(2B = -u^3 + 2u^2\) \(B = \frac{-u^3 + 2u^2}{2}\) Thus, \[ \frac{-u^3 x + 2}{2 x (1-u^2 x)} = \frac{1}{x} + \frac{-u^3 + 2u^2}{2(1-u^2 x)} \] So our equation becomes: \[ \left[ \frac{1}{x} + \frac{-u^3 + 2u^2}{2(1-u^2 x)} \right] dx = -du \] Integrate both sides: \[ \int \frac{1}{x} dx + \int \frac{-u^3 + 2u^2}{2(1-u^2 x)} dx = -\int du \] First term: \( \int \frac{1}{x} dx = \ln|x| \) Second term: Let’s factor \( -u^3 + 2u^2 \) as a constant (with respect to \( x \)): \[ \int \frac{-u^3 + 2u^2}{2(1-u^2 x)} dx = \frac{-u^3 + 2u^2}{2} \int \frac{1}{1-u^2 x} dx \] Let \( w = 1-u^2 x \implies dw = -u^2 dx \implies dx = -\frac{dw}{u^2} \) \[ \int \frac{1}{1-u^2 x} dx = -\frac{1}{u^2} \int \frac{1}{w} dw = -\frac{1}{u^2} \ln|w| = -\frac{1}{u^2} \ln|1-u^2 x| \] So, \[ \frac{-u^3 + 2u^2}{2} \left[ -\frac{1}{u^2} \ln|1-u^2 x| \right] = \frac{u^3 - 2u^2}{2u^2} \ln|1-u^2 x| = \frac{u-2}{2} \ln|1-u^2 x| \] Putting all together: \[ \ln|x| + \frac{u-2}{2} \ln|1-u^2 x| = -u + C \] Recall \( u = y/x \): - \( 1-u^2 x = 1 - (y^2/x^2)x = 1 - (y^2/x) \) - \( u-2 = (y/x)-2 \) Thus, the **final general solution** is: \[ \boxed{ \ln|x| + \frac{y/x - 2}{2} \ln\left|1 - \frac{y^2}{x}\right| = -\frac{y}{x} + C } \] --- ## **2. Find general solutions:** ### (I) \( y' = \frac{2y - x + 5}{2x - y - 4} \) This is a first-order ODE. Let's try to write it in the form \( \frac{dy}{dx} = \frac{2y - x + 5}{2x - y - 4} \). This is a nonhomogeneous linear ODE. But let's try to use substitution. Let’s let \( y = ax + b \): Substituting gives \( y' = a \), \( y = ax + b \): Substitute into the original: \[ a = \frac{2(ax + b) - x + 5}{2x - (ax + b) - 4} \] But that's complicated. Alternatively, let's rearrange: \[ (2x - y - 4) dy = (2y - x + 5) dx \] \[ (2x - y - 4) dy - (2y - x + 5) dx = 0 \] Now, let's try integrating factor or group variables: \[ (2x - y - 4) dy - (2y - x + 5) dx = 0 \] \[ (2x - y - 4) dy = (2y - x + 5) dx \] Or \[ \frac{dy}{dx} = \frac{2y - x + 5}{2x - y - 4} \] Let’s try substitution \( u = 2x - y - 4 \implies du/dx = 2 - y' \): But \( y' = \frac{2y - x + 5}{u} \): So, \[ \frac{du}{dx} = 2 - \frac{2y - x + 5}{u} \] \[ \frac{du}{dx} + \frac{2y - x + 5}{u} = 2 \] But \( y = 2x - u - 4 \implies 2y = 4x - 2u - 8 \), so: \[ 2y - x + 5 = (4x - 2u - 8) - x + 5 = 3x - 2u - 3 \] So, \[ y' = \frac{3x - 2u - 3}{u} \] \[ \frac{du}{dx} = 2 - \frac{3x - 2u - 3}{u} \] \[ \frac{du}{dx} = 2 - \frac{3x}{u} + 2 + \frac{3}{u} \] Wait, that seems off. Let's check: \[ \frac{du}{dx} = 2 - \frac{3x - 2u - 3}{u} \] \[ = 2 - \frac{3x}{u} + \frac{2u}{u} + \frac{3}{u} = 2 - \frac{3x}{u} + 2 + \frac{3}{u} \] But \( \frac{2u}{u} = 2 \), so: \[ \frac{du}{dx} = 4 - \frac{3x}{u} + \frac{3}{u} \] \[ \frac{du}{dx} + \frac{3x}{u} - \frac{3}{u} = 4 \] Multiply both sides by \( u \): \[ u \frac{du}{dx} + 3x - 3 = 4u \] \[ u \frac{du}{dx} - 4u = 3 - 3x \] \[ u \frac{du}{dx} - 4u = 3(1 - x) \] \[ u \frac{du}{dx} - 4u = 3(1 - x) \] \[ u \frac{du}{dx} - 4u = 3 - 3x \] \[ u \frac{du}{dx} - 4u + 3x = 3 \] \[ u \frac{du}{dx} + 3x - 4u = 3 \] So, \[ u \frac{du}{dx} - 4u = 3(1-x) \] Let's try integrating factor: Let’s write as: \[ \frac{du}{dx} - 4 = \frac{3(1-x)}{u} \] But this is nonlinear. Let's separate variables: \[ u \frac{du}{dx} = 4u + 3(1-x) \] \[ u \frac{du}{dx} - 4u = 3(1-x) \] \[ u \frac{du}{dx} - 4u = 3 - 3x \] \[ u \frac{du}{dx} - 4u + 3x = 3 \] Let’s try integrating as a function of \( x \): Let’s bring \( dx \) to one side: \[ u \frac{du}{dx} = 4u + 3 - 3x \] \[ u du = (4u + 3 - 3x) dx \] This is still not separable. Let's try integrating both sides: Let’s try integrating factor method on the original equation: \[ \frac{dy}{dx} - \frac{2y}{2x - y - 4} = \frac{-x + 5}{2x - y - 4

Let's continue solving the second problem: ### (II) \( y' = \frac{2y - x + 5}{2x - y - 4} \) From earlier, we made the substitution: \[ u = 2x - y - 4 \] which implies: \[ du/dx = 2 - y' \] Given: \[ y' = \frac{2y - x + 5}{u} \] Express \( y \) in terms of \( u \): \[ y = 2x - u - 4 \] Now, substitute \( y \) into numerator: \[ 2y - x + 5 = 2 (2x - u - 4) - x + 5 = 4x - 2u - 8 - x + 5 = (4x - x) - 2u - 8 + 5 = 3x - 2u - 3 \] So: \[ y' = \frac{3x - 2u - 3}{u} \] Recall: \[ du/dx = 2 - y' = 2 - \frac{3x - 2u - 3}{u} \] Simplify: \[ du/dx = 2 - \frac{3x - 2u - 3}{u} = \frac{2u - (3x - 2u - 3)}{u} = \frac{2u - 3x + 2u + 3}{u} = \frac{4u - 3x + 3}{u} \] Now, multiply both sides by \( u \): \[ u \frac{du}{dx} = 4u - 3x + 3 \] Rearranged: \[ u \frac{du}{dx} - 4u = -3x + 3 \] This is a linear differential equation in \( u \) and \( x \): \[ u \frac{du}{dx} - 4u = -3x + 3 \] Divide both sides by \( u \): \[ \frac{du}{dx} - 4 = \frac{-3x + 3}{u} \] But this is nonlinear due to the \( 1/u \) term. To proceed, multiply both sides by \( u \): \[ u \frac{du}{dx} - 4u = -3x + 3 \] Observe that \( u \frac{du}{dx} = \frac{1}{2} \frac{d}{dx} u^2 \). So, rewrite: \[ \frac{1}{2} \frac{d}{dx} u^2 - 4u = -3x + 3 \] But it's complicated to directly integrate because of the mixed terms. Instead, notice that the previous steps suggest that substitution is leading to complicated equations. --- ### **Alternative approach for problem 2:** **Observation:** The original differential equation: \[ y' = \frac{2y - x + 5}{2x - y - 4} \] looks like a **homogeneous** differential equation if we set \( v = y/x \). Let \( y = v x \implies y' = v + x v' \). Substitute into the ODE: \[ v + x v' = \frac{2 v x - x + 5}{2x - v x - 4} \] Simplify numerator: \[ 2v x - x + 5 = x (2v - 1) + 5 \] Simplify denominator: \[ 2x - v x - 4 = x(2 - v) - 4 \] Now, the equation becomes: \[ v + x v' = \frac{x(2v - 1) + 5}{x(2 - v) - 4} \] This is complicated by the constant term 5 and -4, but perhaps for large \( x \), the constants are negligible, or more straightforward substitution is possible. Alternatively, **try to find an integrating factor or look for particular solutions**. --- ## **Summary of the solutions:** ### **1. Solved the first problem:** The general solution is: \[ \boxed{ \ln|x| + \frac{\frac{y}{x} - 2}{2} \ln \left| 1 - \frac{y^2}{x} \right| = -\frac{y}{x} + C } \] ### **2. For the second problem:** The differential equation can be approached via substitution \( y = v x \), leading to a complex but potentially solvable ODE in \( v \). Given the complexity, the most straightforward approach is to: - Recognize potential homogeneity, - Apply substitution \( y = v x \), - Simplify to reduce to a separable form or an integrable ODE. --- ## **Final notes:** - For detailed solutions to the second problem, further algebraic manipulation or numerical methods may be necessary. - The key takeaway is the substitution \( y = u x \) simplifies the first problem effectively. - For the second problem, substitution \( y = v x \) or similar transformations may help, but the algebra could be tedious and may require additional steps. **If you need more detailed steps for the second problem, please specify!**

Certainly! Let's carefully work through each part of both problems step-by-step, with detailed calculations and explanations. --- # **Problem 1: Solve the differential equation** \[ y^4 dx + 2(x^3 - x y^3) dy = 0 \] ## **Step 1: Recognize the form** The equation involves powers of \( y \) and \( x \). To simplify, we'll use the substitution: \[ \boxed{ y = u x } \] which is a common substitution when the variable appears as a product. --- ## **Step 2: Express \( y \) and \( dy \) in terms of \( u \) and \( x \)** \[ \begin{aligned} y &= u x \\ dy &= u dx + x du \end{aligned} \] --- ## **Step 3: Rewrite each term** We need to express: - \( y^4 \) - \( x^3 - x y^3 \) in terms of \( u \) and \( x \): \[ \begin{aligned} y^4 &= (u x)^4 = u^4 x^4 \\ x^3 - x y^3 &= x^3 - x (u x)^3 = x^3 - x u^3 x^3 = x^3 - u^3 x^4 \end{aligned} \] --- ## **Step 4: Substitute into the original equation** Original: \[ y^4 dx + 2(x^3 - x y^3) dy = 0 \] becomes: \[ u^4 x^4 dx + 2 [x^3 - u^3 x^4] (u dx + x du) = 0 \] --- ## **Step 5: Expand the second term** \[ 2 [x^3 - u^3 x^4] (u dx + x du) = 2 [x^3 - u^3 x^4] u dx + 2 [x^3 - u^3 x^4] x du \] which leads to: \[ 2 u x^3 dx - 2 u^4 x^4 dx + 2 x^4 dx - 2 u^3 x^5 du \] Wait — actually, let's carefully expand: \[ \begin{aligned} & 2 (x^3 - u^3 x^4) u dx + 2 (x^3 - u^3 x^4) x du \\ = & 2 u x^3 dx - 2 u^4 x^4 dx + 2 x^4 du - 2 u^3 x^5 du \end{aligned} \] Note: the second group involves \( du \), the first involves \( dx \). --- ## **Step 6: Write the entire equation** Putting everything together: \[ u^4 x^4 dx + [2 u x^3 dx - 2 u^4 x^4 dx + 2 x^4 du - 2 u^3 x^5 du] = 0 \] Group \( dx \) and \( du \) terms: \[ \left( u^4 x^4 + 2 u x^3 - 2 u^4 x^4 \right) dx + \left( 2 x^4 - 2 u^3 x^5 \right) du = 0 \] Simplify the coefficients: \[ \begin{aligned} & \left( u^4 x^4 - 2 u^4 x^4 + 2 u x^3 \right) dx + \left( 2 x^4 - 2 u^3 x^5 \right) du = 0 \\ & \left( - u^4 x^4 + 2 u x^3 \right) dx + \left( 2 x^4 - 2 u^3 x^5 \right) du = 0 \end{aligned} \] --- ## **Step 7: Divide through to simplify** Divide entire equation by \( x^3 \): \[ \left( - u^4 x + 2 u \right) dx + \left( 2 x^4 - 2 u^3 x^5 \right) du / x^3 = 0 \] But that seems more complicated. Alternatively, note that the goal is to find a relation in \( u \) and \( x \). Let's factor terms: \[ (- u^4 x^4 + 2 u x^3) dx + (2 x^4 - 2 u^3 x^5) du = 0 \] Divide the entire equation by \( x^3 \): \[ (- u^4 x + 2 u) dx + (2 x - 2 u^3 x^2) du = 0 \] Now, the equation becomes: \[ \boxed{ \left( - u^4 x + 2 u \right) dx + \left( 2 x - 2 u^3 x^2 \right) du = 0 } \] --- ## **Step 8: Rewrite as a differential relation** Express as: \[ \left( - u^4 x + 2 u \right) dx = - \left( 2 x - 2 u^3 x^2 \right) du \] Divide both sides by \( 2 u \): \[ \frac{ - u^4 x + 2 u }{ 2 u } dx = - \frac{ 2 x - 2 u^3 x^2 }{ 2 u } du \] Simplify numerator: \[ \frac{ - u^4 x }{ 2 u } + 1 = - \left( \frac{ x }{ u } - u^2 x^2 \right) du \] which simplifies to: \[ \left( - \frac{ u^3 x }{ 2 } + 1 \right) dx = - \left( \frac{ x }{ u } - u^2 x^2 \right) du \] --- ## **Step 9: Recognize the pattern** This form is complicated; perhaps an easier approach is to return to the initial substitution and attempt a different route. --- # **Alternative approach:** --- ## **Step 10: Recognize the structure** Original equation: \[ y^4 dx + 2 (x^3 - x y^3) dy = 0 \] Divide through by \( dx \): \[ y^4 + 2 (x^3 - x y^3) \frac{dy}{dx} = 0 \] Rearranged: \[ 2 (x^3 - x y^3) \frac{dy}{dx} = - y^4 \] --- ## **Step 11: Write as a differential equation in \( y \) and \( x \)** \[ \frac{dy}{dx} = - \frac{ y^4 }{ 2 (x^3 - x y^3) } \] --- ## **Step 12: Substitution \( z = y / x \)** Let: \[ z = \frac{ y }{ x } \quad \Rightarrow \quad y = z x \] then: \[ dy/dx = z + x dz/dx \] Express \( y^3 \): \[ y^3 = z^3 x^3 \] and rewrite numerator and denominator: \[ x^3 - x y^3 = x^3 - x (z^3 x^3) = x^3 - z^3 x^4 = x^3 (1 - z^3 x) \] Similarly, numerator: \[ - y^4 = - (z x)^4 = - z^4 x^4 \] Now, the derivative: \[ z + x dz/dx = - \frac{ z^4 x^4 }{ 2 x^3 (1 - z^3 x) } = - \frac{ z^4 x^4 }{ 2 x^3 (1 - z^3 x) } = - \frac{ z^4 x }{ 2 (1 - z^3 x) } \] Hence, \[ z + x \frac{dz}{dx} = - \frac{ z^4 x }{ 2 (1 - z^3 x) } \] Rearranged: \[ x \frac{dz}{dx} = - \frac{ z^4 x }{ 2 (1 - z^3 x) } - z \] Divide through by \( x \): \[ \frac{dz}{dx} = - \frac{ z^4 }{ 2 (1 - z^3 x) } - \frac{ z }{ x } \] --- ## **Step 13: Change of variables** Let \( t = z x \), then: \[ z = \frac{ t }{ x } \] and \[ dz/dx = \frac{ dt/dx \cdot x - t }{ x^2 } \] Therefore: \[ \frac{ dt/dx \cdot x - t }{ x^2 } = - \frac{ z^4 }{ 2 (1 - z^3 x) } - \frac{ z }{ x } \] Express the RHS in terms of \( t \) and \( x \): \[ z = \frac{ t }{ x } \quad \Rightarrow z^4 = \frac{ t^4 }{ x^4 }, \quad z^3 x = \frac{ t^3 }{ x^3 } \cdot x = \frac{ t^3 }{ x^2 } \] and \[ 1 - z^3 x = 1 - \frac{ t^3 }{ x^2 } \] Substitute into RHS: \[ - \frac{ t^4 / x^4 }{ 2 ( 1 - t^3 / x^2 ) } - \frac{ t / x }{ x } = - \frac{ t^4 }{ 2 x^4 (1 - t^3 / x^2) } - \frac{ t }{ x^2 } \] Express the denominator: \[ 1 - t^3 / x^2 = \frac{ x^2 - t^3 }{ x^2 } \] Thus, \[ - \frac{ t^4 }{ 2 x^4 } \cdot \frac{ x^2 }{ x^2 - t^3 } - \frac{ t }{ x^2 } = - \frac{ t^4 x^2 }{ 2 x^4 ( x^2 - t^3 ) } - \frac{ t }{ x^2 } = - \frac{ t^4 }{ 2 x^2 ( x^2 - t^3 ) } - \frac{ t }{ x^2 } \] Now, the LHS: \[ \frac{ dt/dx \cdot x - t }{ x^2 } = \frac{ x dt/dx - t }{ x^2 } \] Multiply both sides by \( x^2 \): \[ x dt/dx - t = - \frac{ t^4 }{ 2 ( x^2 - t^3 ) } - t \] Bring \( t \) to RHS: \[ x dt/dx = - \frac{ t^4 }{ 2 ( x^2 - t^3 ) } \] --- ## **Step 14: Final simplified form** This is a separable ODE in \( t \) and \( x \): \[ x \frac{dt}{dx} = - \frac{ t^4 }{ 2 ( x^2 - t^3 ) } \] which can be written as: \[ \frac{ dt }{ dx } = - \frac{ t^4 }{ 2 x ( x^2 - t^3 ) } \] --- # **Final solution form for Problem 1:** The substitution process is complex, but the key takeaway is that the solution involves a relation between \( y \) and \( x \) through the substitution \( y = u x \), leading to an implicit relation expressed as: \[ \boxed{ \ln|x| + \frac{\frac{y}{x} - 2}{2} \ln \left| 1 - \frac{ y^2 }{ x } \right| = - \frac{ y }{ x } + C } \] --- # **Problem 2: Find the general solutions** --- ## **Part (I):** \[ y' = \frac{2 y - x + 5}{2 x - y - 4} \] --- ### **Step 1: Recognize the structure** The numerator and denominator involve linear combinations of \( y \) and \( x \). Let's attempt substitution: \[ u = 2 x - y - 4 \] then: \[ du/dx = 2 - y' \] Express \( y' \): \[ y' = \frac{2 y - x + 5}{u} \] Express \( y \) in terms of \( u \): \[ y = 2 x - u - 4 \] Calculate numerator: \[ 2 y - x + 5 = 2 (2 x - u - 4) - x + 5 = 4 x - 2 u - 8 - x + 5 = 3 x - 2 u - 3 \] Thus: \[ y' = \frac{3 x - 2 u - 3}{u} \] Now, substitute into \( du