The solution of y-2 - 7y-1 - 12y - 0 isy - C-1 e-3x - C-2 e-4xy - C-1xe-3x - C-2 e-4xy - C-1 e-3x - C-2 xe-4xNone-of-these

The correct option is A y = C_1 e^3x + C_2 e^4xThe given equation can be written as (d/dx 3)(dy/dx 4y) = 0 If dy/dx 4y = u then (I) reduces to du/dx 3u = 0 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Solving Linear Differential Equations of First Order

The solution ...

Question

The solution of $y_{2} - 7 y_{1} + 12 y = 0$ is

y = C_{1} e^{3 x} + C_{2} e^{4 x}

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

y = C_{1} x e^{3 x} + C_{2} e^{4 x}

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

y = C_{1} e^{3 x} + C_{2} x e^{4 x}

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

N o n e o f t h e s e

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

Open in App

Solution

The correct option is A $y = C_{1} e^{3 x} + C_{2} e^{4 x}$
The given equation can be written as $(\frac{d}{d x} - 3) (\frac{d y}{d x} - 4 y) = 0$
If $\frac{d y}{d x} - 4 y = u$ then (I) reduces to $\frac{d u}{d x} - 3 u = 0$
$\Rightarrow \frac{d u}{u} = 3 d x \Rightarrow u = C_{1} e^{3 x}$ . Therefore, we have $\frac{d y}{d x} - 4 y = C_{1} e^{3 x}$ which is a linear equation whose I.F. is
$e^{4 x} . S o \frac{d}{d x} (y e^{- 4 x}) = C_{1} e^{- x} \Rightarrow y e^{- 4 x} = - C_{1} e^{- x} + C_{2} \Rightarrow y = C_{1}^{3 x} + C_{2} e^{4 x}$

Suggest Corrections

The solution of y2−7y1+12y=0 is

The solution of $y_{2} - 7 y_{1} + 12 y = 0$ is