Question

Solve the following differential equations:
i) dy/dx = cot(y + x) −1

Answer 1

Solution.  We write the equation as

$$\frac {dy} {dx} =\frac {1} {tan(x+y)} - 1$$

$$\frac {dy} {dx} =\frac {1-tan(x+y} {tan(x+y)}$$

$$tan(x+y)dy=(1-tan(x+y))dx$$

$$(tan(x+y)-1)dx+tan(x+y)dy=0$$

Consider the equation as

$$M(x,y)dx+N(x,y)dy=0.$$

Therefore

$$M(x,y)=tan(x+y)-1$$

$$N(x,y)=tan(x+y)$$

$$\frac {\partial M} {\partial y}= \frac {1} {cos^2 (x+y)}$$

$$\frac {\partial N} {\partial x}= \frac {1} {cos^2 (x+y)}$$ $$\frac {\partial M} {\partial y} = \frac {\partial N} {\partial x}$$

This equation is an equation in full differentials. Let U(x,y) is solution of the equation. Therefore

$$\frac {\partial U} {\partial x}= tan(x+y)-1= \frac {sin(x+y} {cos(x+y)}-1$$

Integrating x we get

$$U(x,y)=-\ln cos(x+y)-x+C(y)$$

where C(y) is function of y. Hence

$$\frac {\partial U} {\partial y}= tan(x+y)+C'(y)= tan(x+y)$$

$$C'(y)=0$$

$$C(y)=C$$

where C is constant. Therefore get

$$U(x,y)=-\ln cos(x+y)-x+C$$

where C is constant.

Answer.

$$U(x,y)=-\ln cos(x+y)-x+C$$

where C is constant.