If y1(x) and y2(x) are two solutions of (d y/d x)+f(x) y=r(x) then y1(x)+y2(x) is solution of:

Question

If y1(x) and y2(x) are two solutions of (d y/d x)+f(x) y=r(x) then y1(x)+y2(x) is solution of: (A) (d y/d x)+f(x) y=0 (B) (d y/d x)+2 f(x) y=r(x) (C)  (d y/d x)+f(x) y=2 r(x) (D) (d y/d x)+2 f(x) y=2 r(x)

Tardigrade Admin · Accepted Answer

(d y1/d x)+f y1=r (d y2/d x)+f y2=r Add (d/d x)(y1+y2)+f(y1+y2)=2 r here (d y/d x)+f(x) y=2 r

Q. If $y_{1} (x)$ and $y_{2} (x)$ are two solutions of $\frac{d y}{d x} + f (x) y = r (x)$ then $y_{1} (x) + y_{2} (x)$ is solution of:

$\frac{d y}{d x} + f (x) y = 0$

$\frac{d y}{d x} + 2 f (x) y = r (x)$

$\frac{d y}{d x} + f (x) y = 2 r (x)$

$\frac{d y}{d x} + 2 f (x) y = 2 r (x)$

$\frac{d y _{1}}{d x} + f y_{1} = r$
$\frac{d y _{2}}{d x} + f y_{2} = r$
Add $\frac{d}{d x} (y_{1} + y_{2}) + f (y_{1} + y_{2}) = 2 r$
here $\frac{d y}{d x} + f (x) y = 2 r$

Q. If y1​(x) and y2​(x) are two solutions of dxdy​+f(x)y=r(x) then y1​(x)+y2​(x) is solution of:

dxdy​+f(x)y=0

dxdy​+2f(x)y=r(x)

dxdy​+f(x)y=2r(x)

dxdy​+2f(x)y=2r(x)