Solve the initial value problem dydx=yx+sec(yx)\frac{dy}{dx} = \frac{y}{x} + \sec(\frac{y}{x})dxdy=xy+sec(xy) with y(1)=0y(1) = 0y(1)=0.
sin(yx)=ln∣x∣\sin(\frac{y}{x}) = \ln|x|sin(xy)=ln∣x∣
sin(yx)=x\sin(\frac{y}{x}) = xsin(xy)=x
cos(yx)=ln∣x∣\cos(\frac{y}{x}) = \ln|x|cos(xy)=ln∣x∣
tan(yx)=ln∣x∣\tan(\frac{y}{x}) = \ln|x|tan(xy)=ln∣x∣