Question

Find the solution of (3x-y+6)dx +(6x-2y-6)dy=0 using Case 1

Answer 1

$$(3x-y+6)dx +(6x-2y-6)dy=0$$

$$\dfrac{dy}{dx}=-\dfrac{(3x-y)+6}{2(3x-y)-6}$$

Let $v=3x-y.$ Then

$$\dfrac{dv}{dx}=3-\dfrac{dy}{dx}=3+\dfrac{v+6}{2v-6}=\dfrac{7v-12}{2v-6}$$

$$\dfrac{2v-6}{7v-12}dv=dx$$

Integrate

$$\int\dfrac{2v-6}{7v-12}dv=\int dx$$

$$\int\dfrac{2v-6}{7v-12}dv=\dfrac{2}{7}\int\dfrac{7v-12}{7v-12}dv-\dfrac{18}{7}\int\dfrac{1}{7v-12}dv$$

$$=\dfrac{2}{7}v-\dfrac{18}{49}\ln(|7v-12|)+C_1$$

$$\dfrac{2}{7}v-\dfrac{18}{49}\ln(|7v-12|)=x-\dfrac{C}{49}$$

$$\dfrac{2}{7}(3x-y)-\dfrac{18}{49}\ln(|21x-7y-12|)=x-\dfrac{C}{49}$$

$$49x-42x+14y+18\ln(|21x-7y-12|)=C$$

The solution is

$$7x+14y+18\ln(|21x-7y-12|)=C$$