TS EAMCET · Maths · Differential Equations
The differential equation for which \(l x^2+m y^2=x+y\) is the general solution is
- A \(\left|\begin{array}{ccc}x^2 & y^2 & x+y \ 2 x & 2 y^{\prime} y & y^{\prime}+1 \ 2 & 2 y y^{\prime \prime} & y^{\prime \prime}\end{array}\right|=0\)
- B \(\left|\begin{array}{ccc}x^2 & y^2 & x+y \ 2 x & 2 y y^{\prime} & 1+y^{\prime} \ 2 & 2\left(y^{\prime 2}+y y^{\prime \prime}\right) & y^{\prime \prime}\end{array}\right|=0\)
- C \(\left|\begin{array}{ccc}x^2 & y^2 & x+y \ 2 x & 2 y y^{\prime} & y+1 \ 2 & 2\left(y^{\prime 2}+y^{\prime} y^{\prime \prime}\right) & y^{\prime \prime}\end{array}\right|=0\)
- D \(\left|\begin{array}{ccc}x^2 & y^2 & x+y \ 2 x & 2 y & 1+y^{\prime} \ 2 & 2 y y^{\prime} y^{\prime \prime} & y^{\prime \prime}\end{array}\right|=0\)
Answer & Solution
Correct Answer
(B) \(\left|\begin{array}{ccc}x^2 & y^2 & x+y \ 2 x & 2 y y^{\prime} & 1+y^{\prime} \ 2 & 2\left(y^{\prime 2}+y y^{\prime \prime}\right) & y^{\prime \prime}\end{array}\right|=0\)
Step-by-step Solution
Detailed explanation
Given equation is \(l x^2+m y^2-x-y=0\) \(\ldots\) (i) On differentiating successively twice we get \(2 l x+2 m y y^{\prime}-1-y^{\prime}=0\) \(\ldots\) (ii) and \(2 l+2 m\left(y y^{\prime \prime}+\left(y^{\prime}\right)^2\right)-y^{\prime \prime}=0\) \(\ldots\) (iii) from Eqs.…
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