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Q.

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 $\frac{d y}{d x} + f (x) y = r (x)$ can be $y =$

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a

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

b

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

c

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

d

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

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{d y}{d x} - f (x) y = 0 \\ \frac{d y}{y} = f (x) d x \\ I n y = \int f (x) d x \end{array}$

$y_{1} (x) = e^{\int f (x) d x}$ 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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If y1x is a solution of the differential equation dydx−fxy=0, then a solution of the differential equation dydx+fxy=rxcan be y=