Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Solution of diffrential equation $\frac{dy}{dx}= \frac{x+2y+1}{x+y+1}$ is

Option 1)
$1/2 ln |y^2-xy-x^2|+ \frac{3}{2\sqrt5} ln |x+y | = C$

Option 2)
$1/2 ln |x^2+y^2+xy|+ {3} ln |x-y | = C$

Option 3)
$1/2 ln |x^2-y^2+xy|+ {4} ln |x^2-xy | = C$

Option 4)
none of these

Answers (1)

best_answer

As we have learned

Equations Reducible to the homogeneous form -

$\frac{dy}{dx}=\frac{ax+by+c}{Ax+By+C}$

- wherein

$x=X+h$

$y=Y+k$

Let x = X+h and y = Y+k then the equation becomes $\frac{dy}{dx}=\frac{(X+h)+ (2Y+k)+1}{(X+h)+(Y+k)+1}= \frac{(x+2Y)+ (h+2k+1)}{(x+y)+ (h+k+1)}$

h and K should be such that

h+2k+1=0 and h+k+1=0

$\Rightarrow h= -1 , k = 0 \Rightarrow x = X-1 and ...y= Y$

Equation becomes $\frac{dy}{dx} = \frac{x+2y}{x+y}$ which is again a homogeneous diffrential equation , so Y = V $\cdot$ X

$\Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}$

$\Rightarrow v + x\frac{dv}{dx}= \frac{x+2vx}{x+vx}\Rightarrow v+x\frac{dv}{dx}= \frac{1+2v}{1+v}$

$\Rightarrow x\frac{dv}{dx}= \frac{1+2v}{1+v}- v \Rightarrow x\frac{dv}{dx} = \frac{1+v-v^2}{1+v}$

$\Rightarrow \frac{1+v }{v^2-v-1}dv+ dx/x = 0$

$\Rightarrow \int \frac{1+v}{v^2-v-1}dx + \int \frac{dx}{x} = C$

$\Rightarrow 1/2\int \frac{(2v-1)+3}{v^2-v-1}dv + \int \frac{dx}{x} = C$

$\Rightarrow 1/2\int \frac{(2v-1)}{v^2-v-1}dv +3/2\int \frac{dv}{v^2-v-1}+ \int \frac{dx}{x} = C$

$\Rightarrow 1/2 ln |v^2-v-1|+ 3/2\sqrt5 ln |\frac{2v-1-\sqrt5}{2v-1+\sqrt5}|+ ln |X|= C$

$\Rightarrow 1/2 ln |y^2/x^2-y/x-1|+ 3/2\sqrt 5 ln |\frac{2y-(1+\sqrt 5)x}{2y- (\sqrt 5 -1)x}|+ ln |X|= C$

$\Rightarrow 1/2 ln|Y^2-XY-X^2|+ 3/2\sqrt 5 ln |\frac{2y-(1+\sqrt 5)x}{2y- (\sqrt 5 -1)x}|= C$

where Y=y and X = x +1

on putting these values , we are not getting (A)(B)(C)

Option 1)

$1/2 ln |y^2-xy-x^2|+ \frac{3}{2\sqrt5} ln |x+y | = C$

Option 2)

$1/2 ln |x^2+y^2+xy|+ {3} ln |x-y | = C$

Option 3)

$1/2 ln |x^2-y^2+xy|+ {4} ln |x^2-xy | = C$

Option 4)

none of these

Posted by

gaurav

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Advanced 2024: Chemical engineering cut-offs at IITs; know opening, closing ranks

anam-khan

IIT JEE Mains Result 2024 Session 2: How is overall merit list prepared?

anam-khan

JEE Mains Session 2 result 2024 soon; List of top NITs in India

Latest Question