JEE Main & Advanced Mathematics Differential Equations Solution by Inspection

If we can write the differential equation in the form \[f({{f}_{1}}(x,\,y))d({{f}_{1}}(x,\,y))+\varphi ({{f}_{2}}(x,\,y))d({{f}_{2}}(x,\,y))+......=0\], then each term can be easily integrated separately. For this the following results must be memorized.

(i) \[d(x+y)=dx+dy\]

(ii) \[d(xy)=xdy+ydx\]

(iii) \[d\left( \frac{x}{y} \right)=\frac{ydx-xdy}{{{y}^{2}}}\]                 

(iv) \[d\left( \frac{y}{x} \right)=\frac{xdy-ydx}{{{x}^{2}}}\]

(v) \[d\,\left( \frac{{{x}^{2}}}{y} \right)=\frac{2xydx-{{x}^{2}}dy}{{{y}^{2}}}\]    

(vi) \[d\left( \frac{{{y}^{2}}}{x} \right)=\frac{2xydy-{{y}^{2}}dx}{{{x}^{2}}}\]

(vii) \[d\left( \frac{{{x}^{2}}}{{{y}^{2}}} \right)=\frac{2x{{y}^{2}}dx-2{{x}^{2}}ydy}{{{y}^{4}}}\]

(viii) \[d\,\left( \frac{{{y}^{2}}}{{{x}^{2}}} \right)=\frac{2{{x}^{2}}ydy-2x{{y}^{2}}dx}{{{x}^{4}}}\]

(ix) \[d\,\left( {{\tan }^{-1}}\frac{x}{y} \right)=\frac{ydx-xdy}{{{x}^{2}}+{{y}^{2}}}\]

(x) \[d\left( {{\tan }^{-1}}\frac{y}{x} \right)=\frac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}\]

(xi) \[d[\ln (xy)]=\frac{xdy+ydx}{xy}\]   

(xii) \[d\left( \ln \left( \frac{x}{y} \right) \right)=\frac{ydx-xdy}{xy}\]

(xiii) \[d\,\left[ \frac{1}{2}\ln ({{x}^{2}}+{{y}^{2}}) \right]=\frac{xdx+ydy}{{{x}^{2}}+{{y}^{2}}}\]

(xiv) \[d\left[ \ln \left( \frac{y}{x} \right) \right]=\frac{xdy-ydx}{xy}\]

(xv) \[d\text{ }\left( -\frac{1}{xy} \right)=\frac{xdy+ydx}{{{x}^{2}}{{y}^{2}}}\]    

(xvi) \[d\text{ }\left( \frac{{{e}^{x}}}{y} \right)=\frac{y{{e}^{x}}dx-{{e}^{x}}dy}{{{y}^{2}}}\]               

(xvii) \[d\left( \frac{{{e}^{y}}}{x} \right)=\frac{x{{e}^{y}}dy-{{e}^{y}}dx}{{{x}^{2}}}\]

(xviii) \[d({{x}^{m}}{{y}^{n}})={{x}^{m-1}}{{y}^{n-1}}(mydx+nxdy)\]

(xix) \[d\text{ }\left( \sqrt{{{x}^{2}}+{{y}^{2}}} \right)=\frac{xdx+y\,dy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}\]

(xx) \[d\text{ }\left( \frac{1}{2}\log \frac{x+y}{x-y} \right)=\frac{x\,dy-y\,dx}{{{x}^{2}}-{{y}^{2}}}\]

(xxi) \[\frac{d{{[f(x,\,y)]}^{1-n}}}{1-n}=\frac{{f}'(x,\,y)}{{{(f(x,\,y))}^{n}}}\]