MCQ
The solution of the differential equation ${(x + y)^2}\frac{{dy}}{{dx}} = {a^2}$ is
- A${(x + y)^2} = \frac{{{a^2}}}{2}x + c$
- B${(x + y)^2} = {a^2}x + c$
- C${(x + y)^2} = 2{a^2}x + c$
- ✓None of these
✓
Answer
Correct option: D.
None of these
d
(d) Put $x + y = v$ ==> $1 + \frac{{dy}}{{dx}} = \frac{{dv}}{{dx}}$ ==> $\frac{{dy}}{{dx}} = \frac{{dv}}{{dx}} - 1$
(d) Put $x + y = v$ ==> $1 + \frac{{dy}}{{dx}} = \frac{{dv}}{{dx}}$ ==> $\frac{{dy}}{{dx}} = \frac{{dv}}{{dx}} - 1$
${v^2}\left( {\frac{{dv}}{{dx}} - 1} \right) = {a^2}$
==> $\frac{{dv}}{{dx}} = \frac{{{a^2}}}{{{v^2}}} + 1 = \frac{{{a^2} + {v^2}}}{{{v^2}}}$ ==> $\frac{{{v^2}}}{{{a^2} + {v^2}}}dv = dx$
==> $\left( {1 - \frac{{{a^2}}}{{{a^2} + {v^2}}}} \right)dv = dx$ ==> $v - a{\tan ^{ - 1}}\frac{v}{a} = x + c$
==> $y = a{\tan ^{ - 1}}\left( {\frac{{x + y}}{a}} \right)+ c.$
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