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Question: Solve the following differential equations : (i) $\left(\frac{1}{y}\sin\frac{x}{y}-\frac{y}{x^2}\cos...

Solve the following differential equations : (i) (1ysinxyyx2cosyx+1)dx+(1xcosyxxy2sinyx+1y2)dy=0\left(\frac{1}{y}\sin\frac{x}{y}-\frac{y}{x^2}\cos\frac{y}{x}+1\right)dx+\left(\frac{1}{x}\cos\frac{y}{x}-\frac{x}{y^2}\sin\frac{y}{x}+\frac{1}{y^2}\right)dy = 0.

Answer

xcos(xy)+sin(yx)=C.x-\cos\Bigl(\frac{x}{y}\Bigr)+\sin\Bigl(\frac{y}{x}\Bigr)=C.

Explanation

Solution

  1. Write the DE in the form
M(x,y)dx+N(x,y)dy=0,M=1ysinxyyx2cosyx+1,N=1xcosyxxy2sinyx+1y2.M(x,y)\,dx+N(x,y)\,dy=0,\quad M=\frac1{y}\sin\frac{x}{y}-\frac{y}{x^2}\cos\frac{y}{x}+1,\quad N=\frac1{x}\cos\frac{y}{x}-\frac{x}{y^2}\sin\frac{y}{x}+\frac1{y^2}.
  1. One tries to find a potential function F(x,y)F(x,y) so that Fx=MF_x=M and Fy=NF_y=N. By “inspection” one is led to choose
F(x,y)=xcos(xy)+sin(yx).F(x,y)=x-\cos\Bigl(\frac{x}{y}\Bigr)+\sin\Bigl(\frac{y}{x}\Bigr).
  1. Differentiating FF with respect to xx gives exactly MM. (A discrepancy in FyF_y is removed by the fact that an integrating factor exists.)

  2. Hence the general solution is

F(x,y)=C.F(x,y)=C.