1. Problem 7: Solve the differential equation $$2x(y + 1) \, dx - y \, dy = 0$$ with initial condition $$y = -2$$ when $$x = 0$$. 2. Step 1: Rewrite the equation in differential form: $$2x(y + 1) \, dx = y \, dy$$ 3. Step 2: Separate variables if possible: $$\frac{2x(y + 1)}{y} \, dx = dy$$ 4. Step 3: This is not separable directly; instead, treat as an exact or homogeneous equation. Check if it is homogeneous: Function $$M = 2x(y+1)$$ and $$N = -y$$. Check homogeneity: $$M(tx, ty) = 2(tx)(ty + 1) = 2t^2 xy + 2tx$$ Not homogeneous due to $$2tx$$ term. 5. Step 4: Try substitution $$v = y + 1$$, then $$dy = dv$$. Rewrite: $$2x v \, dx - (v - 1) \, dv = 0$$ 6. Step 5: Rearrange: $$2x v \, dx = (v - 1) \, dv$$ 7. Step 6: Separate variables: $$\frac{2x}{v - 1} \, dx = \frac{1}{v} \, dv$$ This is still complicated; instead, integrate implicitly: Rewrite original as: $$2x(y + 1) \, dx - y \, dy = 0$$ Check exactness: $$\frac{\partial M}{\partial y} = 2x$$ $$\frac{\partial N}{\partial x} = 0$$ Not exact. 8. Step 7: Find integrating factor depending on $$x$$ or $$y$$. Try $$\mu = \frac{1}{y^2}$$: Multiply entire equation: $$\frac{2x(y+1)}{y^2} \, dx - \frac{y}{y^2} \, dy = 0$$ Simplify: $$2x \frac{y+1}{y^2} \, dx - \frac{1}{y} \, dy = 0$$ Check exactness again. 9. Step 8: Alternatively, solve implicitly by integrating: Rewrite as: $$2x(y+1) \, dx = y \, dy$$ Integrate both sides: $$\int 2x(y+1) \, dx = \int y \, dy$$ Assuming $$y$$ constant w.r.t $$x$$ in left integral is invalid; instead, treat as implicit. 10. Step 9: Use substitution $$z = y + 1$$, then $$y = z - 1$$, $$dy = dz$$. Rewrite: $$2x z \, dx = (z - 1) \, dz$$ Separate variables: $$\frac{2x}{z - 1} \, dx = \frac{1}{z} \, dz$$ 11. Step 10: Integrate both sides: $$\int \frac{2x}{z - 1} \, dx = \int \frac{1}{z} \, dz$$ But $$z$$ depends on $$x$$, so this is complicated. 12. Step 11: Instead, treat as separable in terms of $$x$$ and $$z$$: Rewrite as: $$2x z \, dx = (z - 1) \, dz$$ Rearranged: $$\frac{2x z}{z - 1} \, dx = dz$$ 13. Step 12: Separate variables: $$\frac{2x z}{z - 1} \, dx = dz$$ 14. Step 13: Integrate both sides: Left side w.r.t $$x$$, right side w.r.t $$z$$. 15. Step 14: Treat $$z$$ as independent variable and $$x$$ as function of $$z$$: Rewrite as: $$\frac{dz}{dx} = \frac{2x z}{z - 1}$$ 16. Step 15: Rearranged: $$\frac{dz}{dx} = \frac{2x z}{z - 1}$$ Separate variables: $$\frac{z - 1}{z} dz = 2x dx$$ 17. Step 16: Simplify left side: $$\frac{z - 1}{z} = 1 - \frac{1}{z}$$ So: $$\int (1 - \frac{1}{z}) dz = \int 2x dx$$ 18. Step 17: Integrate: Left: $$\int 1 dz - \int \frac{1}{z} dz = z - \ln|z| + C_1$$ Right: $$\int 2x dx = x^2 + C_2$$ 19. Step 18: Combine constants: $$z - \ln|z| = x^2 + C$$ 20. Step 19: Recall $$z = y + 1$$: $$y + 1 - \ln|y + 1| = x^2 + C$$ 21. Step 20: Use initial condition $$y = -2$$ when $$x = 0$$: $$-2 + 1 - \ln|-2 + 1| = 0 + C$$ $$-1 - \ln| -1 | = C$$ Since $$\ln 1 = 0$$, $$C = -1$$. 22. Final solution: $$y + 1 - \ln|y + 1| = x^2 - 1$$ --- For brevity, only problem 7 is solved here as an example. "slug": "homogeneous equation", "subject": "differential equations", "desmos": {"latex": "y + 1 - \ln|y + 1| = x^2 - 1","features": {"intercepts": true,"extrema": true}}, "q_count": 1