The solution of y 2 − 7 y 1 + 12 y = 0 is Options: A . &nbsp y = C 1 e 3 x + C 2 e 4 x B . &nbsp y = C 1 x e 3 x + C 2 e 4 x C . &nbsp y = C 1 e 3 x + C 2 x e 4 x D . &nbsp N o n e o f t h e s e

Answer: Option A : A The given equation can be written as ( d d x − 3 ) ( d y d x − 4 y ) = 0 . . . . ( i ) If d y d x − 4 y = u then (I) reduces to d u d x − 3 u = 0 ⇒ d u u = 3 d x ⇒ u = C 1 e 3 x . Therefore, we have d y d x − 4 y = C 1 e 3 x which is a linear equation whose I.F.is e − 4 x . So d d x ( y e − 4 x ) = C 1 e − x ⇒ y e − 4 x = − C 1 e − x + C 2 ⇒ y = C 1 e 3 x + C 2 e 4 x

The solution of y2−7y1+12y=0 is

Discussion Forum : Differential Equations

Question -

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

Options:

A .

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

B .

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

C .

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

D .

N o n e o f t h e s e

Answer: Option A
:
A
The given equation can be written as

(\frac{d}{d x} - 3) (\frac{d y}{d x} - 4 y) = 0 . . . . (i)

\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}

. So

\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} e^{3 x} + C_{2} e^{4 x}

Was this answer helpful ?

Next Question

Discussion Forum : Differential Equations

Submit Your Solution hear: