Question

The general solution of the differential equation $\frac{dy}{dx}=\frac{(x+y)+(x+y-1)\ln (x+y)}{\ln (x+y)}$ is (where $C$ is a constant of integration)

• A. $\ln \frac{x+y}{|1+\ln (x+y\left)\right|}=x+C$
• B. $\ln \frac{xy}{(1+\ln (xy\left)\right)}=x^2+C$
• C. $\ln \frac{|x-y|}{\ln (1+\ln |x-y|)}=y+C$
• D. $\ln \frac{x^2+y^2}{(1+\ln (x^2+y^2\left)\right)}=x^2+C$

Answer 1

The correct option is A $\ln \frac{x+y}{|1+\ln (x+y\left)\right|}=x+C$

The substitution $x+y=v$ will reduce this differential equation to the following Variable Separable form:
$\frac{dv}{dx}-1=\frac{v+(v-1)\ln v}{\ln v}$
$=(v-1)+\frac{v}{\ln v}$
$\Rightarrow \frac{dv}{dx}=v+\frac{v}{\ln v}$
$\Rightarrow \frac{\ln v}{v(1+\ln v)}dv=dx\Rightarrow \int \frac{\ln v}{v(1+\ln v)}dv=\int dx$
To evaluate the integral on the L.H.S, we use the substitution $(1+\ln v)=t$ which gives $\frac{1}{v}dv=dt$.
$\int \frac{t-1}{t}dt=\int dx$
$\Rightarrow t-\ln |t|=x+C$
$\Rightarrow (1+\ln v)-\ln |1+\ln v|=x+C$
$\Rightarrow (1+\ln (x+y\left)\right)-\ln |1+\ln (x+y)|=x+C$
$\Rightarrow \ln \frac{x+y}{|1+\ln (x+y\left)\right|}=x+C$