Differentiation (Part-38) | S N De

Find the particular solutions of the following equations :  

$~26.~~\frac{dy}{dx}=\frac{3x+2y}{2x-3y};\\~~\text{given}~~~y=0,~\text{when}~x=1.$

Sol. $~~~~\frac{dy}{dx}=\frac{3x+2y}{2x-3y} \\ \text{or,}~~ \frac{dy}{dx}=\frac{3+2(y/x)}{2-3(y/x)}\rightarrow(1)$

Let $~~~~y=vx ~(\Rightarrow v=y/x)\\ \therefore \frac{dy}{dx}=v+x\frac{dv}{dx}\rightarrow(2)$

So, from $~(1),(2)~$ we get,

$~~v+x\frac{dv}{dx}=\frac{3+2v}{2-3v} \\ \text{or,}~~ x\frac{dv}{dx}=\frac{3+2v}{2-3v}-v \\ \text{or,}~~x\frac{dv}{dx}=\frac{3+2v-v(2-3v)}{2-3v} \\ \text{or,}~~x\frac{dv}{dx}=\frac{3+2v-2v+3v^2}{2-3v} \\ \text{or,}~~x~\frac{dv}{dx}=\frac{3+3v^2}{2-3v} \\ \text{or,}~~ \frac{2-3v}{3(1+v^2)}~dv=\frac{dx}{x} \\ \text{or,}~~\frac{3v-2}{3(v^2+1)}~dv=-\frac{dx}{x} \\ \text{or,}~~ \frac{3(v-2/3)}{3(v^2+1)}~dv=\frac{dx}{x} \\ \text{or,}~~ \frac 12\left(\frac{2v-4/3}{v^2+1}\right)~dv=-\frac 1x~dx \\ \therefore \frac 12 \int{\frac{2v}{v^2+1}~dv}-\frac 12 \int{\frac{4/3}{v^2+1}~dv}=-\int{\frac{dx}{x}} \\ \text{or,}~~ \frac 12 \log|v^2+1|-\frac 12 \cdot \frac 43 \tan^{-1}(v)=-\log|x|+c \\ \text{or,}~~ \frac 12 \log\left|\frac{y^2}{x^2}+1\right|-\frac 23 \tan^{-1}(y/x)+\log|x|=c \rightarrow(3)$

Now, $\\~~\text{given}~~~y=0,~\text{when}~~x=1.$

So, from $~(3),~$ we get,

$~~\frac 12\log 1-\frac 23 \tan^{-1}(0)+\log 1=c \\ \text{or,}~~c=0.$

Hence, putting the value of $\,c\,$ in $\,(3)\,$, we get,

$~~~~\frac 12 \log\left|\frac{x^2+y^2}{x^2}\right|-\frac 23\tan^{-1}(y/x)+\log|x|=0 \\ \text{or,}~~ \log\left|\frac{x^2+y^2}{x^2}\right|-\frac 43\tan^{-1}(y/x)+2\log|x|=0  \\ \text{or,}~~\log\left|\frac{x^2+y^2}{x^2}\right|-\frac 43\tan^{-1}(y/x)+\log|x|^2=0  \\ \text{or,}~~ \log\left|\frac{x^2+y^2}{x^2} \cdot x^2\right|-\frac 43\tan^{-1}(y/x)=0 \\ \text{or,}~~ 3\log|x^2+y^2|-4\tan^{-1}(y/x)=0 \\ \therefore 4\tan^{-1}(y/x)-3\log(x^2+y^2)=0~~\text{(ans.)}$

Note : $~~c \rightarrow \text{constant of integration.}$

$~27.~~\frac{dy}{dx}=\frac{4x+y}{x+y}; \\~~~\text{given}~~y=4,~~\text{when}~~x=1.$

Sol. $~~\frac{dy}{dx}=\frac{4x+y}{x+y} \\ \text{or,}~~ \frac{dy}{dx}=\frac{4+(y/x)}{1+(y/x)}  \rightarrow(1)$

Let $~~~~y=vx \\ \therefore \frac{dy}{dx}=v+x\frac{dv}{dx}\rightarrow(2)$

So, from $~(1),(2)~$ we get,

$~~~v+x\frac{dv}{dx}=\frac{4+v}{1+v} \\ \text{or,}~~ x\frac{dv}{dx}=\frac{4+v}{1+v}-v \\ \text{or,}~~x\frac{dv}{dx}=\frac{4+v-v(1+v)}{1+v} \\ \text{or,}~~x\frac{dv}{dx}=\frac{4+v-v-v^2}{1+v} \\ \text{or,}~~x\frac{dv}{dx}=\frac{4-v^2}{1+v} \\ \text{or,}~~ \frac{v+1}{4-v^2}~dv=\frac 1x~dx \\ \text{or,}~~ -\frac 12\left(\frac{-2v-2}{4-v^2}\right)~dv=\frac 1x~dx \\ \text{or,}~~-\frac 12 \int{\frac{-2v~dv}{4-v^2}}+\int{\frac{dv}{2^2-v^2}}=\int{\frac{dx}{x}} \\ \text{or,}~~ -\frac 12\log|4-v^2|+\frac{1}{2 \times 2} \log\left|\frac{2+v}{2-v}\right|\\~~=\log|x|+\log c_1 \\ \text{or,}~~ -\frac 12 \log\left|4-\frac{y^2}{x^2}\right|+\frac 14 \log\left|\frac{2+(y/x)}{2-(y/x)}\right|\\~~=\log|x|+\log c_1 \\ \text{or,}~~ -2\log\left|\frac{4x^2-y^2}{x^2}\right|+\log\left|\frac{2x+y}{2x-y}\right|\\~~=4\log |x|+4\log c_1 \\ \text{or,}~~ \log \left|\frac{x^4}{(4x^2-y^2)^2} \times \frac{2x+y}{2x-y}\right|\\~~=\log |x|^4+\log c_1^4 \\ \text{or,}~~ \log \left|\frac{x^4}{(4x^2-y^2)^2} \times \frac{2x+y}{2x-y}\right|-\log |x|^4=\log c_1^4  \\ \text{or,}~~ \log \left|\frac{x^4}{(4x^2-y^2)^2} \times \frac{2x+y}{2x-y} \cdot \frac{1}{x^4}\right|=\log c_1^4 \\ \text{or,}~~ \log \left|\frac{1}{(4x^2-y^2)^2} \times \frac{2x+y}{2x-y}\right|=\log c \\ \text{or,}~~\left|\frac{1}{(4x^2-y^2)^2} \times \frac{2x+y}{2x-y}\right|= c \\ \text{or,}~~ \frac{(2x+y)^2}{(4x^2-y^2)^4(2x-y)^2}=c^2 \rightarrow(3)$

Now, $~~\text{given}~~y=4,~~\text{when}~~x=1.$

So, by $~(3),~$ we get,

$~~c^2=\frac{(2 \cdot 1+4)^2}{(4 \cdot 1-4^2)^4(2 \cdot 1-4)^2}=\frac{1}{2304}.$

Putting the value of $~c^2,~$ we get from $~(3),$

$~~~\frac{(2x+y)^2}{(4x^2-y^2)^4(2x-y)^2}=\frac{1}{2304} \\ \text{or,}~~  2304(2x+y)^2\\~~~=(4x^2-y^2)^4(2x-y)^2~~\text{(ans.)}$

Note : $~~\log c_1 \rightarrow \text{constant of integration.},~~c_1^4=c.$

$~28.~~\frac{dy}{dx}-\frac yx+\csc(y/x)=0~~~~[\text{CBSE}-2009]\\~~\text{given}~~~y=0~~\text{when}~~x=1.$

Sol.  Here by $~\csc(y/x),~$ we mean cosec$(y/x).$

Now, $~\frac{dy}{dx}-\frac yx+\csc(y/x)=0 \rightarrow(1)$

Let $~~~~y=vx \\ \therefore \frac{dy}{dx}=v+x\frac{dv}{dx}\rightarrow(2)$

So, from $~(1),(2)~$ we get,

$~~v+x~\frac{dv}{dx}-v+\csc v=0 \\ \text{or,}~~ x~\frac{dv}{dx}=-\csc v \\ \text{or,}~~ \int{\frac{dv}{\csc v}}=-\int{\frac{dx}{x}} \\ \text{or,}~~ \int{\sin v ~dv}=-\log|x|+\log c_1 \\ \text{or,}~~ -\cos|v|=-\log|x|+\log c_1 \\ \text{or,}~~ -\cos\left|\frac yx\right|+\log|x|=\log c_1 \rightarrow(2)$

For $~~x=1,~y=0,~~$ we get by $~(2),$

$~~~-\cos 0+ \log 1=\log c_1 \\ \text{or,}~~ -1+0=\log c_1 \\ \text{or,}~~ \log c_1=-1.$

Putting the value of $~\log c_1,~$ we get from $~(2),$

$~~-\cos|y/x|+\log|x|=-1 \\ \text{or,}~~ \log|x|+1=\cos(y/x)~~\text{(ans.)}$

Note : $~~\log c_1 \rightarrow \text{constant of integration.}$