The differential equation whose solution is y - Ae3x + Be-3x is given by

Question

The differential equation whose solution is y - Ae3x + Be-3x is given by (A) y2 - 3y1 + 3y = 0 (B) xy2+3y1-xy+x2+3=0 (C) y2-9y=0 (D) (y1)3-3y(xy2-3y)=0

Tardigrade Admin · Accepted Answer

We have, y = Ae3x + Be-3x Differentiating w.r.t. x, we get y1 = 3Ae3x -3 Be-3x Again differentiating w.r.t. x, we get y2=9 Ae3x+9 Be-3x=9(Ae3x+Be-3x)=9y ⇒ y2-9y=0

Q. The differential equation whose solution is $y - A e^{3 x} + B e^{- 3 x}$ is given by

$y_{2} - 3 y_{1} + 3 y = 0$

$x y_{2} + 3 y_{1} - x y + x^{2} + 3 = 0$

$y_{2} - 9 y = 0$

$(y_{1})^{3} - 3 y (x y_{2} - 3 y) = 0$

We have, $y = A e^{3 x} + B e^{- 3 x}$
Differentiating w.r.t. $x$ , we get
$y_{1} = 3 A e^{3 x} - 3 B e^{- 3 x}$
Again differentiating w.r.t. $x$ , we get
$y_{2} = 9 A e^{3 x} + 9 B e^{- 3 x} = 9 (A e^{3 x} + B e^{- 3 x}) = 9 y$
$\Rightarrow y_{2} - 9 y = 0$

Q. The differential equation whose solution is y−Ae3x+Be−3x is given by

y2​−3y1​+3y=0

xy2​+3y1​−xy+x2+3=0

y2​−9y=0

(y1​)3−3y(xy2​−3y)=0