JEE Main & Advanced Mathematics Differential Equations Variable Separable Type Differential Equation

(1) Equations in variable separable form : If the differential equation of the form \[{{f}_{1}}(x)dx={{f}_{2}}(y)dy\]        .....(i)

where \[{{f}_{1}}\] and \[{{f}_{2}}\] being functions of \[x\] and \[y\] only. Then we say that the variables are separable in the differential equation.

Thus, integrating both sides of (i), we get its solution as \[\int_{{}}^{{}}{{{f}_{1}}(x)dx}=\int_{{}}^{{}}{{{f}_{2}}(y)dy}+c\], where \[c\] is an arbitrary constant.

There is no need of introducing arbitrary constants to both sides as they can be combined together to give just one.

(2) Equations reducible to variable separable form

(i) Differential equations of the form \[\frac{dy}{dx}=f(ax+by+c)\] can be reduced to variable separable form by the substitution  \[ax+by+c=Z\].

\[\therefore \] \[a+b\frac{dy}{dx}=\frac{dZ}{dx}\];  \[\therefore \] \[\left( \frac{dZ}{dx}-a \right)\frac{1}{b}=f(Z)\]\[\Rightarrow \]\[\frac{dZ}{dx}=a+bf(Z)\].

This is variable separable form.

(ii) Differential equation of the form

\[\frac{dy}{dx}=\frac{ax+by+c}{Ax+By+C}\], where \[\frac{a}{A}=\frac{b}{B}=K\] (say)

\[\therefore \] \[\frac{dy}{dx}=\frac{K(Ax+By)+c}{Ax+By+C}\]

Put \[Ax+By=Z\Rightarrow \] \[A+B\frac{dy}{dx}=\frac{dZ}{dx}\]

\[\therefore \]\[\left[ \frac{dZ}{dx}-A \right]\frac{1}{B}=\frac{KZ+c}{Z+C}\]\[\Rightarrow \] \[\frac{dZ}{dx}=A+B\frac{KZ+c}{Z+C}\]

This is variable separable form and can be solved.