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  • Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 66 Subquestion (i) textbook solution.

Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 66 Subquestion (i) textbook solution.

Answers (1)

Answer          : \log \left|x^{2}+x y+y^{2}\right|=2 \sqrt{3} \tan ^{-1}\left(\frac{x+2 y}{\sqrt{3} x}\right)+c

Hint                :  You must know the rules of solving differential equation and integration

Given              :  (x-y) \frac{d y}{d x}=x+2 y

Solution          :

STEP : 1

\frac{dy}{dx}=\frac{x+2y}{x-y}

STEP : 2

\begin{aligned} &\text { Put } f(x)=\frac{d y}{d x} \quad, \text { find } f(\lambda x, \lambda y) \\ &\frac{d y}{d x}=\frac{x+2 y}{x-y} \\ &\text { Put } f(x, y)=\frac{x+2 y}{x-y} \\ &\text { Find } f(\lambda x, \lambda y) \end{aligned}

\begin{aligned} f(\lambda x, \lambda y) &=\frac{\lambda x+2(\lambda y)}{\lambda x-\lambda y} \\ &=\frac{\lambda(x+2 y)}{\lambda(x-y)} \\ &=\frac{x+2 y}{x * y} \end{aligned}                        .....(i)

                     =f(x,y)

STEP : 3

\text {Put}\; y=vx

\begin{aligned} &\frac{d y}{d x}=\frac{d(v x)}{d x} \\ &\text { So } \frac{d y}{d x}=\frac{d v}{d x} \times x+v \frac{d x}{d x} \\ &\frac{d y}{d x}=x \frac{d v}{d x}+v \\ &\text { Put } \frac{d y}{d x} \text { and } \frac{y}{x} \text { in }(i) \end{aligned}

\begin{aligned} &\frac{d v}{d x} x+v=\frac{x+2 v x}{x-v x} \\ &\frac{d v}{d x} x+v=\frac{x(1+2 v)}{x(1-v)} \\ &\frac{d v}{d x} x=\frac{(1+2 v-v(1-v))}{1-v} \end{aligned}

         \begin{gathered} \frac{d v}{d x} x=\frac{1+2 v-v+v^{2}}{1-v} \\ \frac{d v}{d x} x=\frac{v^{2}+v+1}{1-v} \\ \frac{d v}{d x} x=-\left(\frac{v^{2}+v+1}{v-1}\right) \end{gathered}

\begin{aligned} &d v\left(\frac{v-1}{v^{2}+v+1}\right)=-\frac{d x}{x} \\ &\int \frac{(v-1)}{v^{2}+v+1} d v=-\int \frac{d x}{x} \\ &\int \frac{(v-1)}{v^{2}+v+1} d v=-\log |x|+C \end{aligned}

\begin{aligned} &\text { Use } v^{2}+v+1=v^{2}+\frac{1}{2} \times 2 v+\frac{\left(1^{2}\right)}{\left(2^{2}\right)}+1-\frac{\left(1^{2}\right)}{2^{2}} \\ &\left(v+\frac{1}{2}\right)^{2}+1-\frac{1}{4} \\ &\left(v+\frac{1}{2}\right)^{2}+\frac{3}{4} \end{aligned}

\begin{aligned} &\text { Put } v^{2}+v+1=\left(v+\frac{1}{2}\right)^{2}+\frac{3}{4} \\ &v-1=v+\frac{1}{2}-\frac{1}{2}-1=\left(v+\frac{1}{2}\right)-\frac{3}{2} \end{aligned}

\begin{aligned} &\int \frac{\left(v+\frac{1}{2}\right)-\frac{3}{2}}{\left(v+\frac{1}{2}\right)^{2}+\frac{3}{4}} d v=-\log |x|+C \\ &\int \frac{v+\frac{1}{2}}{\left(v+\frac{1}{4}\right)^{2}+\frac{3}{4}} d v-\frac{3}{2} \int \frac{1}{\left(v+\frac{1}{2}\right)^{2}+\frac{3}{4}} d v=-\log |x|+C \end{aligned}

So, our equation become

I_{1}-\frac{3}{2} I_{2}=-\log |x|+c                                    ....(ii)

Solving I_{1},

\begin{aligned} &I_{1}=\int \frac{v+\frac{1}{2}}{\left(v+\frac{1}{2}\right)^{2}+\frac{3}{4}} d v \\ &\text { Put }\left(v+\frac{1}{2}\right)^{2}+\frac{3}{4}=t \end{aligned}

Diff.w.r.t.v

\begin{aligned} &\frac{d\left(\left(v+\frac{1}{2}\right)^{2}+\frac{3}{4}\right)}{d v}=\frac{d t}{d v} \\ &2\left(v+\frac{1}{2}\right)=\frac{d t}{d v} \\ &d v=\frac{d t}{2\left(v+\frac{1}{2}\right)} \end{aligned}

Put value of v and dv in I_{1}

\begin{aligned} &I_{1}=\int \frac{v+\frac{1}{2}}{t} \times \frac{d t}{2\left(v+\frac{1}{2}\right)} \\ &=\frac{1}{2} \int \frac{d t}{t}=\frac{1}{2} \log |t| \\ &=\frac{1}{2} \log \left|\left(v+\frac{1}{2}\right)^{2}+\frac{3}{2}\right| \\ &\frac{1}{2} \log \left|v^{2}+v+1\right| \end{aligned}

Solving I_{1}

\begin{aligned} &I_{2}=\int \frac{d v}{\left(v+\frac{1}{2}\right)^{2}+\frac{3}{4}} \\ &\quad=\int \frac{d v}{\left(v+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} \\ &\text { Using } \int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c \end{aligned}

\begin{aligned} &\text { Where } \mathrm{x}=v+\frac{1}{2} \text { and } a=\frac{\sqrt{3}}{2} \\ &=\frac{1}{\frac{\sqrt{3}}{2}} \tan ^{-1} \frac{\left(v+\frac{1}{2}\right)}{\frac{\sqrt{3}}{2}} \\ &=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 v+1}{\sqrt{3}}\right) \end{aligned}

From (ii)

\begin{aligned} &I_{1}-\frac{3}{2} I_{2}=-\log |x|+C \\ &\frac{1}{2} \log \left|v^{2}+v+1\right|-\frac{3}{2} \times \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 v+1}{\sqrt{3}}\right)=-\log |x|+C \\ &\frac{1}{2} \log \left|v^{2}+v+1\right|-\sqrt{3} \tan ^{-1} \frac{2 v+1}{\sqrt{3}}=-\log |x|+C \end{aligned}

Replace v by \frac{y}{x}

\begin{aligned} &\frac{1}{2} \log \left|\left(\frac{y}{x}\right)^{2}+\frac{y}{x}+1\right|-\sqrt{3} \tan ^{-1}\left(\frac{2 \frac{y}{x}+1}{\sqrt{3}}\right)=-\log |x|+C \\ &\frac{1}{2} \log \left|\frac{y^{2}}{x^{2}}+\frac{y}{x}+1\right|+\log |x|=\sqrt{3} \tan ^{-1}\left(\frac{2 y+x}{\sqrt{3} x}\right)+C \end{aligned}

Multiply both side by 2

\begin{aligned} &\log \left|\frac{y^{2}}{x^{2}}+\frac{y}{x}+1\right|+\log |x|^{2}=2 \sqrt{3} \tan ^{-1}\left(\frac{2 y+x}{\sqrt{3} x}\right)+2 C \\ &\text { Put } 2 c=c \\ &\log \left(\frac{x^{2} y^{2}}{x^{2}}+\frac{x^{2} y}{x}+x^{2}\right)=2 \sqrt{3} \tan ^{-1}\left(\frac{x+2 y}{\sqrt{3} x}\right)+C \\ &\log \left|y^{2}+x y+x^{2}\right|=2 \sqrt{3} \tan ^{-1}\left(\frac{x+2 y}{\sqrt{3} x}\right)+C \end{aligned}

 

 

 

Posted by

infoexpert23

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