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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y_{2} - 9 y = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $y_{2} - 9 y = 0$
$y = A e^{3 x} + B e^{- 3 x}$ ...(1)
Differentiating w.r.t. $y$ we get
$y_{1} = 3 A e^{3 x} - 3 B e^{- 3 x}$ ...(2)
Again differentiating w.r.t $y$ we get
$y_{2} = 9 A e^{3 x} + 9 B e^{- 3 x}$ ...(3)
Therefore, from (1) and (3), we get
$y_{2} - 9 y = 0$

Suggest Corrections

Similar questions

Q. The solution of the differential equation

(x^{2} - y x^{2}) \frac{d y}{d x} + y^{2} + x y^{2} = 0

Q. If

y + 3 x = 0

is the equation of a chord of the circle,

x^{2} + y^{2} - 30 x = 0

, then the equation of the circle with this chord as diameter is

Q. The equation of the circle which cuts orthogonally each of the three circles given below:

x^{2} + y^{2} - 2 x + 3 y - 7 = 0

x^{2} + y^{2} + 5 x - 5 y + 9 = 0

and

x^{2} + y^{2} + 7 x - 9 y + 29 = 0

Find the equation of circle circumscribing the quadrilateral formed by four lines $3 x + 4 y - 5 = 0$ , $4 x - 3 y - 5 = 0$ , $3 x + 4 y + 5 = 0$ and $4 x - 3 y + 5 = 0$

Q. The equation of the circle whose diameter is the common chord of the circles

x^{2} + y^{2} + 2 x + 3 y + 2 = 0

x^{2} + y^{2} + 2 x - 3 y - 4 = 0

Join BYJU'S Learning Program

The differential equation whose solution is y=Ae3x+Be−3x is given by:

The correct option is B y2−9y=0y=Ae3x+Be−3x ...(1)Differentiating w.r.t. y we gety1=3Ae3x−3Be−3x ...(2)Again differentiating w.r.t y we gety2=9Ae3x+9Be−3x ...(3)Therefore, from (1) and (3), we gety2−9y=0

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

The correct option is B $y_{2} - 9 y = 0$
$y = A e^{3 x} + B e^{- 3 x}$ ...(1)
Differentiating w.r.t. $y$ we get
$y_{1} = 3 A e^{3 x} - 3 B e^{- 3 x}$ ...(2)
Again differentiating w.r.t $y$ we get
$y_{2} = 9 A e^{3 x} + 9 B e^{- 3 x}$ ...(3)
Therefore, from (1) and (3), we get
$y_{2} - 9 y = 0$