If y 1 ( x ) is a solution of the differential equation d y d x − f ( x ) y = 0 , then a solution of the differential equation d y d x + f ( x ) y = r ( x ) Options: A . &nbsp 1 y 1 ( x ) ∫ r ( x ) y 1 ( x ) d x B . &nbsp y 1 ( x ) ∫ r ( x ) y 1 ( x ) d x C . &nbsp ∫ r ( x ) y 1 ( x ) d x D . &nbsp N o n e o f t h e s e

Answer: Option A : A d y d x − f ( x ) . y = 0 d y y = f ( x ) d x ln y = ∫ f ( x ) d x y 1 ( x ) = e ∫ f ( x ) d x Then for given equation I.F = e ∫ f ( x ) d x Hence Solution y . y 1 ( x ) = ∫ r ( x ) . y 1 ( x ) d x y = 1 y 1 ( x ) ∫ r ( x ) . y 1 ( x ) d x

If y1(x) is a solution of the differential equation dydx−f(x)y=0,&

Discussion Forum : Differential Equations

Question -

If $y_{1} (x)$ is a solution of the differential equation $\frac{d y}{d x} - f (x) y = 0$ , then a solution of the differential equation $\frac{d y}{d x} + f (x) y = r (x)$

Options:

A .

\frac{1}{y_{1} (x)} \int r (x) y_{1} (x) d x

B .

y_{1} (x) \int \frac{r (x)}{y_{1} (x)} d x

C .

\int r (x) y_{1} (x) d x

D .

N o n e o f t h e s e

Answer: Option A
:
A

\frac{d y}{d x} - f (x) . y = 0 \frac{d y}{y} = f (x) d x

y = \int f (x) d x

y_{1} (x) = e^{\int f (x) d x}

Then for given equation I.F =

e^{\int f (x) d x}

Hence Solution

y . y_{1} (x) = \int r (x) . y_{1} (x) d x

y = \frac{1}{y_{1} (x)} \int r (x) . y_{1} (x) d x

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