If the solution of the differential equation (d y/d x)=(x3 + x y2/y3 - y x2) is yk-xk=2x2y2+λ (where, λ is an arbitrary constant), then the value of k is

Question

If the solution of the differential equation (d y/d x)=(x3 + x y2/y3 - y x2) is yk-xk=2x2y2+λ (where, λ is an arbitrary constant), then the value of k is (A) 2 (B) 4 (C) 1 (D) (3/2)

Tardigrade Admin · Accepted Answer

We have, (d y/d x)=(x3 + x y2/y3 - y x2) Now, (y d y/x d x)=(x2 + y2/y2 - x2) Let, y2=Y;x2=X ⇒ (y d y/x d x)=(d Y/d X) Hence, the equation is (d Y/d X)=(X + Y/Y - X) ⇒ YdY-XdX=XdY+YdX On integrating we get (Y2/2)-(X2/2)= displaystyle ∫ d (X Y) = X Y + c or Y2-X2=2XY+λ ⇒ y4-x4=2x2y2+λ

Q. If the solution of the differential equation $\frac{d y}{d x} = \frac{x ^{3} + x y ^{2}}{y ^{3} - y x ^{2}}$ is $y^{k} - x^{k} = 2 x^{2} y^{2} + λ$ (where, $λ$ is an arbitrary constant), then the value of $k$ is

$2$

$4$

$1$

$\frac{3}{2}$

Q. If the solution of the differential equation dxdy​=y3−yx2x3+xy2​ is yk−xk=2x2y2+λ (where, λ is an arbitrary constant), then the value of k is

2

4

1

23​

We have, dxdy​=y3−yx2x3+xy2​ Now, xdxydy​=y2−x2x2+y2​ Let, y2=Y;x2=X ⇒xdxydy​=dXdY​ Hence, the equation is dXdY​=Y−XX+Y​ ⇒YdY−XdX=XdY+YdX On integrating we get 2Y2​−2X2​=∫d(XY)=XY+c or Y2−X2=2XY+λ ⇒y4−x4=2x2y2+λ

Q. If the solution of the differential equation $\frac{d y}{d x} = \frac{x ^{3} + x y ^{2}}{y ^{3} - y x ^{2}}$ is $y^{k} - x^{k} = 2 x^{2} y^{2} + λ$ (where, $λ$ is an arbitrary constant), then the value of $k$ is

$2$

$4$

$1$

$\frac{3}{2}$