Advertisements
Advertisements
प्रश्न
If y = 500e7x + 600e-7x, then show that y2 – 49y = 0.
Advertisements
उत्तर
y = 500e7x + 600e-7x
`y_1 = "dy"/"dx" = 500 "d"/"dx" (e^(7x)) + 600 "d"/"dx" (e^(-7x))`
`= 500 (7e^(7x)) + 600(- 7e^(-7x))`
`y_2 = ("d"^2"y")/"dx"^2 = 500xx7 "d"/"dx" (e^(7x)) + 600(-7) "d"/"dx" (e^(-7x))`
`= 500 xx 7(7e^(7x)) + 600 xx (-7)(-7) e^(-7x)`
`= 500 xx 49e^7x + 600 xx 49e^(-7x)`
`y_2 = 49 [500 r^(7x) + 600e^(-7x)]` = 49y
(or) y2 – 49y = 0
APPEARS IN
संबंधित प्रश्न
Differentiate the following with respect to x.
x3 ex
Differentiate the following with respect to x.
`(sqrtx + 1/sqrtx)^2`
Differentiate the following with respect to x.
`e^x/(1 + e^x)`
Differentiate the following with respect to x.
`sqrt(1 + x^2)`
Find `"dy"/"dx"` for the following function
xy – tan(xy)
If 4x + 3y = log(4x – 3y), then find `"dy"/"dx"`
If xm . yn = (x + y)m+n, then show that `"dy"/"dx" = y/x`
Find `"dy"/"dx"` of the following function:
x = log t, y = sin t
If xy . yx , then prove that `"dy"/"dx" = y/x((x log y - y)/(y log x - x))`
If xy2 = 1, then prove that `2 "dy"/"dx" + y^3`= 0
