Let's solve the **first question** step by step: ### **1. By considering \( y = u^x \), solve the following equation:** \[ y^4 dx + 2(x^3 - x y^3) dy = 0 \] --- #### **Step 1: Substitute \( y = u^x \)** Let \( y = u^x \Rightarrow \ln y = x \ln u \). Now, compute \( dy \): \[ y = u^x \implies \frac{dy}{dx} = x u^{x-1} \frac{du}{dx} + u^x \ln u \] But since we are dealing with differentials, let's use \( dy = \frac{\partial y}{\partial u} du + \frac{\partial y}{\partial x} dx \). But, generally, we can introduce a substitution to simplify the equation. --- #### **Step 2: Express Everything in Terms of \(u\) and \(x\)** Let’s try a direct substitution for \( y = u^x \): - \( y^4 = (u^x)^4 = u^{4x} \) - \( y^3 = (u^x)^3 = u^{3x} \) Now, express \( dy \): \[ y = u^x \implies dy = x u^{x-1} du + u^x \ln u dx \] But, since we want to substitute for \(dy\), let's keep this in mind. --- #### **Step 3: Substitute All Terms Into the ODE** The original equation is: \[ y^4 dx + 2(x^3 - x y^3) dy = 0 \] Substitute \( y^4 \) and \( y^3 \): \[ u^{4x} dx + 2[x^3 - x u^{3x}] dy = 0 \] Let's isolate \( dy \): \[ u^{4x} dx = -2[x^3 - x u^{3x}] dy \] \[ \frac{u^{4x}}{2[x^3 - x u^{3x}]} dx = -dy \] \[ dy = -\frac{u^{4x}}{2[x^3 - x u^{3x}]} dx \] --- #### **Step 4: Solve the Differential Equation** Now, integrate both sides with respect to \( x \): \[ \int dy = -\int \frac{u^{4x}}{2[x^3 - x u^{3x}]} dx \] But this is problematic since \(u\) can be a function of \(x\). If we treat \(u\) as a constant (if \(y = u^x\) and \(u\) is a parameter), then integrate. But more likely, what is meant is to use the substitution \(y = u^x\), then try to reduce the order or make the equation separable. --- #### **Step 5: Try Another Substitution** Let’s try expressing \( y \) in terms of \( x \): Let’s try a substitution: \( y = v x \), or \( y = v \) (if \(u\) is a function of \(x\)), but since the question asks for \( y = u^x \), let's proceed. Let’s try making the equation separable by dividing by \( dx \): \[ y^4 + 2(x^3 - x y^3) \frac{dy}{dx} = 0 \] \[ 2(x^3 - x y^3) \frac{dy}{dx} = -y^4 \] \[ \frac{dy}{dx} = -\frac{y^4}{2(x^3 - x y^3)} \] Now, let’s try \( y = u x \), but since the question says \( y = u^x \), let's go back. Try \( y = u^x \): \[ \frac{dy}{dx} = u^x \ln u \] Substitute in: \[ (u^x)^4 + 2(x^3 - x (u^x)^3) u^x \ln u = 0 \] \[ u^{4x} + 2(x^3 - x u^{3x}) u^x \ln u = 0 \] \[ u^{4x} + 2x^3 u^x \ln u - 2x u^{4x} \ln u = 0 \] Group \(u^{4x}\) terms: \[ u^{4x} - 2x u^{4x} \ln u + 2x^3 u^x \ln u = 0 \] \[ u^{4x} (1 - 2x \ln u) + 2x^3 u^x \ln u = 0 \] \[ u^{4x} (1 - 2x \ln u) = -2x^3 u^x \ln u \] \[ u^{4x} (1 - 2x \ln u) = -2x^3 u^x \ln u \] If \(u \neq 0\), divide both sides by \(u^x\): \[ u^{3x} (1 - 2x \ln u) = -2x^3 \ln u \] \[ u^{3x} (1 - 2x \ln u) + 2x^3 \ln u = 0 \] This is a transcendental equation for \(u\) in terms of \(x\). One possible solution is if \(\ln u = 0 \implies u = 1\): If \( u = 1 \): \[ u^{3x} (1 - 2x \cdot 0) + 2x^3 \cdot 0 = 1 \cdot 1 = 1 \] So for \(u = 1\), the equation is not satisfied unless more simplification is done. Alternatively, if \(1 - 2x \ln u = 0\): \[ 1 - 2x \ln u = 0 \implies \ln u = \frac{1}{2x} \implies u = e^{1/(2x)} \] Therefore, the general solution is: \[ y = u^x = \left(e^{1/(2x)}\right)^x = e^{1/2} \] So for all \(x\), \(y = e^{1/2}\) is a constant solution. Alternatively, the implicit solution is: \[ u^{3x} (1 - 2x \ln u) + 2x^3 \ln u = 0 \] --- ## **Final Answer:** ### **Explicit solution:** \[ \boxed{y = e^{1/2}} \] ### **Implicit general solution:** \[ \boxed{y^{3} (1 - 2x \ln y) + 2x^3 \ln y = 0} \] --- **Explanation:** We used the substitution \(y = u^x\), derived the resulting equation, and solved for \(u\), leading to the final solution above.