1. **Problem Statement:** Find the isogonal trajectories that cut at an angle of 45 degrees to the family of lines passing through the origin. 2. **Given Family:** The family of lines passing through the origin can be written as $$y = mx$$ where $m$ is the slope. 3. **Slope of Given Family:** The slope of these lines is $m = \frac{dy}{dx} = \tan \theta$ where $\theta$ is the angle the line makes with the x-axis. 4. **Isogonal Trajectories:** These are curves that intersect the given family at a constant angle $\alpha = 45^\circ$. 5. **Formula for Isogonal Trajectories:** If the slope of the given family is $m$, and the slope of the isogonal trajectories is $m_1$, then the angle between them satisfies: $$\tan \alpha = \left| \frac{m_1 - m}{1 + m m_1} \right|$$ 6. **Substitute $\alpha = 45^\circ$:** Since $\tan 45^\circ = 1$, we have $$1 = \left| \frac{m_1 - m}{1 + m m_1} \right|$$ 7. **Solve for $m_1$:** This gives two cases: Case 1: $$\frac{m_1 - m}{1 + m m_1} = 1 \implies m_1 - m = 1 + m m_1$$ $$m_1 - m m_1 = 1 + m \implies m_1(1 - m) = 1 + m \implies m_1 = \frac{1 + m}{1 - m}$$ Case 2: $$\frac{m_1 - m}{1 + m m_1} = -1 \implies m_1 - m = -1 - m m_1$$ $$m_1 + m m_1 = m - 1 \implies m_1(1 + m) = m - 1 \implies m_1 = \frac{m - 1}{1 + m}$$ 8. **Recall $m = \frac{dy}{dx}$ for the given family:** For the family $y = mx$, the slope $m = \frac{y}{x}$. 9. **Replace $m$ by $\frac{y}{x}$ in $m_1$:** Case 1: $$m_1 = \frac{1 + \frac{y}{x}}{1 - \frac{y}{x}} = \frac{\frac{x + y}{x}}{\frac{x - y}{x}} = \frac{x + y}{x - y}$$ Case 2: $$m_1 = \frac{\frac{y}{x} - 1}{1 + \frac{y}{x}} = \frac{\frac{y - x}{x}}{\frac{x + y}{x}} = \frac{y - x}{x + y}$$ 10. **Differential equations for isogonal trajectories:** Case 1: $$\frac{dy}{dx} = \frac{x + y}{x - y}$$ Case 2: $$\frac{dy}{dx} = \frac{y - x}{x + y}$$ 11. **Solve Case 1:** Rewrite as: $$ (x - y) dy = (x + y) dx $$ Separate variables or use substitution $v = \frac{y}{x}$: $$ y = vx \implies dy = v dx + x dv $$ Substitute: $$(x - vx)(v dx + x dv) = (x + vx) dx$$ $$x(1 - v)(v dx + x dv) = x(1 + v) dx$$ Divide both sides by $x$: $$(1 - v)(v dx + x dv) = (1 + v) dx$$ Expand: $$(1 - v) v dx + (1 - v) x dv = (1 + v) dx$$ Bring terms involving $dx$ to one side: $$(1 - v) x dv = (1 + v) dx - (1 - v) v dx = \left[(1 + v) - v(1 - v)\right] dx$$ Simplify bracket: $$(1 + v) - v + v^2 = 1 + v^2$$ So: $$(1 - v) x dv = (1 + v^2) dx$$ Rewrite: $$\frac{dv}{dx} = \frac{1 + v^2}{x(1 - v)}$$ Separate variables: $$\frac{1 - v}{1 + v^2} dv = \frac{1}{x} dx$$ Integrate both sides: $$\int \frac{1 - v}{1 + v^2} dv = \int \frac{1}{x} dx$$ Split integral: $$\int \frac{1}{1 + v^2} dv - \int \frac{v}{1 + v^2} dv = \ln |x| + C$$ Integrate: $$\arctan v - \frac{1}{2} \ln(1 + v^2) = \ln |x| + C$$ Replace $v = \frac{y}{x}$: $$\arctan \left( \frac{y}{x} \right) - \frac{1}{2} \ln \left(1 + \left( \frac{y}{x} \right)^2 \right) = \ln |x| + C$$ Simplify logarithm: $$\ln \left(1 + \frac{y^2}{x^2} \right) = \ln \left( \frac{x^2 + y^2}{x^2} \right) = \ln (x^2 + y^2) - 2 \ln |x|$$ Rewrite equation: $$\arctan \left( \frac{y}{x} \right) - \frac{1}{2} \left[ \ln (x^2 + y^2) - 2 \ln |x| \right] = \ln |x| + C$$ Simplify: $$\arctan \left( \frac{y}{x} \right) - \frac{1}{2} \ln (x^2 + y^2) + \ln |x| = \ln |x| + C$$ Cancel $\ln |x|$ on both sides: $$\arctan \left( \frac{y}{x} \right) - \frac{1}{2} \ln (x^2 + y^2) = C$$ This implicit equation represents the isogonal trajectories for Case 1. 12. **Solve Case 2:** $$\frac{dy}{dx} = \frac{y - x}{x + y}$$ Use substitution $v = \frac{y}{x}$, $y = vx$, $dy = v dx + x dv$: $$v dx + x dv = \frac{v x - x}{x + v x} dx = \frac{x(v - 1)}{x(1 + v)} dx = \frac{v - 1}{1 + v} dx$$ Rearranged: $$x dv = \frac{v - 1}{1 + v} dx - v dx = \left( \frac{v - 1}{1 + v} - v \right) dx$$ Simplify bracket: $$\frac{v - 1}{1 + v} - v = \frac{v - 1 - v(1 + v)}{1 + v} = \frac{v - 1 - v - v^2}{1 + v} = \frac{-1 - v^2}{1 + v}$$ So: $$x \frac{dv}{dx} = \frac{-1 - v^2}{1 + v}$$ Rewrite: $$\frac{dv}{dx} = \frac{-1 - v^2}{x(1 + v)}$$ Separate variables: $$\frac{1 + v}{1 + v^2} dv = - \frac{1}{x} dx$$ Integrate both sides: $$\int \frac{1 + v}{1 + v^2} dv = - \int \frac{1}{x} dx$$ Split integral: $$\int \frac{1}{1 + v^2} dv + \int \frac{v}{1 + v^2} dv = - \ln |x| + C$$ Integrate: $$\arctan v + \frac{1}{2} \ln (1 + v^2) = - \ln |x| + C$$ Replace $v = \frac{y}{x}$: $$\arctan \left( \frac{y}{x} \right) + \frac{1}{2} \ln \left(1 + \left( \frac{y}{x} \right)^2 \right) = - \ln |x| + C$$ Simplify logarithm as before: $$\arctan \left( \frac{y}{x} \right) + \frac{1}{2} \left[ \ln (x^2 + y^2) - 2 \ln |x| \right] = - \ln |x| + C$$ Expand: $$\arctan \left( \frac{y}{x} \right) + \frac{1}{2} \ln (x^2 + y^2) - \ln |x| = - \ln |x| + C$$ Add $\ln |x|$ to both sides: $$\arctan \left( \frac{y}{x} \right) + \frac{1}{2} \ln (x^2 + y^2) = C$$ This implicit equation represents the isogonal trajectories for Case 2. **Final answer:** The isogonal trajectories cutting at 45 degrees to the family of lines through the origin satisfy either $$\arctan \left( \frac{y}{x} \right) - \frac{1}{2} \ln (x^2 + y^2) = C$$ or $$\arctan \left( \frac{y}{x} \right) + \frac{1}{2} \ln (x^2 + y^2) = C$$