The integral curves of dx/a = dy/b = dz is
Let us find the integral curves of dxa=dyb=dz.\frac{dx}{a} = \frac{dy}{b} = dz.adx=bdy=dz. It is equivalent to dz=dxadz=\frac{dx}{a}dz=adx and dz=dyb.dz= \frac{dy}{b}.dz=bdy. Then∫dz=∫dxa\int dz=\int\frac{dx}{a}∫dz=∫adx and ∫dz=∫dyb.\int dz= \int\frac{dy}{b}.∫dz=∫bdy. It follows that z=xa+C1z=\frac{x}{a} +C_1z=ax+C1 and z=yb+C2.z= \frac{y}{b}+C_2.z=by+C2. Therefore, z=xa+C1=yb+C2z=\frac{x}{a} +C_1=\frac{y}{b}+C_2z=ax+C1=by+C2 is the required integral curves.
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