Question

solve differential equations (2x+y-1)dy = (x-2y+5)dx​

Answer 1

Step-by-step explanation:

We have,

$(2x + y - 1)dy = (x - 2y + 5)dx$

$\implies \frac{dy}{dx} = \frac{(x - 2y + 5)}{(2x + y - 1)}$

$Let x= u + h \:\:and \:\:y= v + k ....(where \:\: h \:\: and \:\: k are constants$

$\implies \frac{dv}{du} = \frac{((u + h) - 2(v + k)+ 5)}{(2(u + h)+( v + k) - 1)}$

$\implies \frac{dv}{du} = \frac{(u- 2v + h - 2k+ 5)}{(2u + v +2 h+ k - 1)}$

Now, putting

$h - 2k + 5 = 0 \: \: and \: \: 2h + k - 1 = 0$

$h = - \frac{3}{5} \: \: and \: \: k = \frac{11}{5}$

Now, the required differential equation becomes,

$\implies \frac{dv}{du} = \frac{(u- 2v )}{(2u + v )}$

$\implies (2u + v)dv = (u- 2v )du$

$\implies 2udv + vdv = udu- 2v du$

$\implies 2udv + 2vdu = udu- v dv$

$\implies 2(udv + vdu )= udu- v dv$

$\implies 2d(uv) = udu- v dv$

Integrating both sides

$\implies \int2d(uv) = \int \: udu- \int v dv$

$\implies 2uv = u- v + C$

$\implies 2(x - \frac{3}{5})( y + \frac{11}{5} ) = x - \frac{ 3}{5} - y - \frac{11}{5} + C$