recategorized by
349 views

1 Answer

0 votes
0 votes
This question can be solved by using homogeneous equation concept.

Rearranging the equation,   $\frac{dx}{dy}=\frac{x^{2}-y^{2}}{2xy}$    On RHS ,divide the numerator and denominator by $y^{2}$ and let   $x=vy$.

So,  $\frac{dx}{dy}=v+y\frac{dv}{dy}$  $=$  $\frac{v^{2}-1}{2v}$

$\Rightarrow$   $y\frac{dv}{dy}=\frac{-\left ( 1+v^{2} \right )}{2v}$

$\Rightarrow$    $\int \frac{2v}{1+v^{2}}dv=-\int \frac{dy}{y}$

$\Rightarrow$     $ln_{e}\left ( 1+v^{2} \right )=-ln_{e}\left ( y \right )+ln_{e}\left ( c \right )$

$\Rightarrow$     $ln_{e}\left ( \left ( 1+v^{2} \right )\times y \right )=ln_{e}\left ( c \right )$

$\Rightarrow$     $x^{2}+y^{2}=cy$  ---- $\left \{ x=vy \right \}$  substitution in the previous step

Using the given conditions i.e.  $y=10$    for    $x=0$,  we get $c=10$

$\therefore$    $x^{2}+y^{2}=10y$

Rearranging and simplifying the above equation, we get   $x^{2}+\left ( y-5 \right )^{2}=25$

Option A is the correct answer.

Related questions

0 votes
0 votes
1 answer
1
gatecse asked Sep 18, 2019
260 views
The general solution of the differential equation $2y{y}'-x=0$ is (assuming $C$ as an arbitrary constant of integration)$x^{2}-y^{2}=C$$2x^{2}-y^{2}=C$$2y^{2}-x^{2}=C$$x^...
0 votes
0 votes
1 answer
2
gatecse asked Sep 18, 2019
253 views
The general solution of the differential equation $x+y-x{y}'=0$ is (assuming $C$ as an arbitrary constant of integration)$y=x(\log x+C)$$x=y(\log y+C)$$y=x(\log y+C)$$y=y...
1 votes
1 votes
1 answer
3
gatecse asked Sep 18, 2019
389 views
Let $I=\int(\sin\:x-\cos\:x)(\sin\:x+\cos\:x)^{3}dx$ and $K$ be a constant of integration. Then the value of $I$ is$(\sin\:x+\cos\:x)^{4}+K$$(\sin\:x+\cos\:x)^{2}+K$$-\fr...
0 votes
0 votes
1 answer
4
gatecse asked Sep 18, 2019
397 views
The Taylor series expansion of $f(x)=\ln(1+x^{2})$ about $x=0$ is$\sum_{n=1}^{\infty}(-1)^{n}\frac{x^{n}}{n}$$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{2n}}{n}$$\sum_{n=1}^{\...