Question

Consider the differential equation $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$. If two particular solutions of the given equation u(x) and v(x) are known, find the general solution of the same equation in term of u(x) and v(x).

• A. $y=u\left(x\right)+K\left(u\right(x)-v(x\left)\right)$ where K is any constant
• B. $y=u\left(x\right)v\left(x\right)+K\left(u\right(x)+v(x\left)\right)$ where K is any constant
• C. $y=u\left(x\right)+K\left(v\right(x\left)\right)$ where K is any constant
• D. none of these

Answer 1

Answer: (A)

$y=u\left(x\right)+K\left(u\right(x)-v(x\left)\right)$ where K is any constant

$y=u\left(x\right)+K\left(u\right(x)-v(x\left)\right)$ where $K$ is any constant

$\frac{du}{dx}+Pu=Q;\frac{dv}{dx}+Pv=Q$

$\Rightarrow \frac{d}{dx}(u-v)=-P(u-v)$

$\Rightarrow \frac{d(u-v)}{u-v}=-Pdx$

$\Rightarrow ln(u-v)=-\int Pdx$

$\frac{dy}{dx}+Py=Qx$

I.F. $=\frac{1}{u-v}$

$y.\frac{1}{u-v}=\int \frac{Q}{u-v}+c$

$\Rightarrow \frac{u}{u-v}=\int \frac{Q}{u-v}+c^{\prime }$ [u satifies it]

$\therefore \frac{y}{u-v}=\frac{u}{u-v}+k$

$\Rightarrow y=u+k(u-v)$ ..... (1)