Solution of the differential equation [ydx + ( x + x^2y )dy = 0] is

Question

Solution of the differential equation $$ydx + \left( {x + {x^2}y} \right)dy = 0$$ is

Accepted Answer

$$ydx + \left( {x + {x^2}y} \right)dy = 0$$

$$ \Rightarrow {{dx} \over {dy}} =  - {x \over y} - {x^2}$$

$$ \Rightarrow {{dx} \over {dy}} + {x \over y} =  - {x^2},$$

It is Bernoullis form. Divide by ${x^2}$

$${x^{ - 2}}{{dx} \over {dy}} + {x^{ - 1}}\left( {{1 \over y}} \right) =  - 1.$$

put $${x^{ - 1}} = t,\,\, - {x^{ - 2}}{{dx} \over {dy}} = {{dt} \over {dy}}$$

We get, $$ - {{dt} \over {dy}} + t\left( {{1 \over y}} \right) =  - 1$$

$$ \Rightarrow {{dt} \over {dy}} - \left( {{1 \over y}} \right)t = 1$$

It is linear in $t.$

Integrating factor

$$ = {e^{\int { - {1 \over y}dy} }} = {e^{ - \log y}} = {y^{ - 1}}$$

$$\therefore$$ Solution is

$$t\left( {{y^{ - 1}}} \right) = \int {\left( {{y^{ - 1}}} \right)} dy + c$$

$$ \Rightarrow {1 \over x}.{1 \over y} = \log y + c$$

$$ \Rightarrow \log y - {1 \over {xy}} = c$$

Solutions · 1