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give answer in 2 step with explanation at the end of each step and final answer at the end:(8) Show that (y* + 2zy)dz — 2°dy = 0 is a homogeneous D.E. and then solve. Pap

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give answer in 2 step with explanation at the end of each step and final answer at the end:Uploaded Image(8) Show that (y* + 2zy)dz — 2°dy = 0 is a homogeneous D.E. and then solve. Pap

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Asked by: Pagal
Created at: 2025-07-27 05:43:34

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Let's solve the problem **step by step** as requested: --- ### **Step 1: Show that the equation is homogeneous** Given Differential Equation: \[(y^2 + 2xy)dx - x^2dy = 0\] Rewrite as: \[\frac{dy}{dx} = \frac{y^2 + 2xy}{x^2}\] A differential equation is **homogeneous** if the right-hand side is a function of \( \frac{y}{x} \) alone. Let’s check: \[ \frac{y^2 + 2xy}{x^2} = \frac{y^2}{x^2} + \frac{2xy}{x^2} = \left(\frac{y}{x}\right)^2 + 2\left(\frac{y}{x}\right) \] This is a function of \( \frac{y}{x} \) only. **Explanation:** The equation is homogeneous because the right-hand side can be written as a function of \( \frac{y}{x} \) alone. --- ### **Step 2: Solve the homogeneous differential equation** Let \( v = \frac{y}{x} \implies y = vx \). Then, \( \frac{dy}{dx} = v + x\frac{dv}{dx} \). Substitute into the equation: \[ v + x\frac{dv}{dx} = v^2 + 2v \] \[ x\frac{dv}{dx} = v^2 + 2v - v \] \[ x\frac{dv}{dx} = v^2 + v \] \[ \frac{dv}{v^2 + v} = \frac{dx}{x} \] Now, split the left side by partial fractions: \[ \frac{1}{v^2 + v} = \frac{1}{v(v+1)} = \frac{A}{v} + \frac{B}{v+1} \] Solve for A and B: \[ 1 = A(v+1) + Bv \implies \begin{cases} A + B = 0 \\ A = 1 \end{cases} \implies A = 1, B = -1 \] So, \[ \frac{1}{v^2 + v} = \frac{1}{v} - \frac{1}{v+1} \] Now integrate both sides: \[ \int \left(\frac{1}{v} - \frac{1}{v+1}\right) dv = \int \frac{dx}{x} \] \[ \ln|v| - \ln|v+1| = \ln|x| + C \] \[ \ln\left|\frac{v}{v+1}\right| = \ln|x| + C \] \[ \left|\frac{v}{v+1}\right| = K|x| \quad \text{where } K = e^C \] Recall \( v = \frac{y}{x} \): \[ \left|\frac{y/x}{y/x + 1}\right| = K|x| \] \[ \left|\frac{y}{y + x}\right| = K|x| \] --- ## **Final Answer:** **(a) The given equation is homogeneous.** **(b) The solution is:** \[ \boxed{\left|\frac{y}{y + x}\right| = K|x|} \] Where \( K \) is an arbitrary constant.

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This is AI generated Response.
Asked by: Pagal
Created at: 2025-07-27 05:44:56

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The differential equation is homogeneous because the right-hand side can be expressed as a function of \( \frac{y}{x} \). The solution to the equation is: \[ \boxed{\left|\frac{y}{y + x}\right| = C|x|} \] where \( C \) is an arbitrary constant.

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