Advertisements
Advertisements
प्रश्न
If y = 3 cos (log x) + 4 sin (log x), show that x2y2 + xy1 + y = 0.
Advertisements
उत्तर
Given, y = 3 cos (log x) + 4 sin (log x) ...(1)
Differentiating both sides with respect to x,
`dy/dx = 3 d/dx cos (log x) + 4 d/dx sin (log x)`
= `3 [- sin (log x)] d/dx (log x) + 4 cos (log x) d/dx (log x)`
= `-3 sin (log x) xx 1/x + 4 cos (log x) xx 1/x`
Multiplying both sides by x,
`x dy/dx` = −3 sin (log x) + 4 cos (log x)
Differentiating both sides again with respect to x,
`x d/dx (dy/dx) + dy/dx * d/dx (x) = - 3 cos (log x) d/dx (log x) - 4 sin (log x) d/dx (log x)`
`x (d^2 y)/dx^2 + 1 * dy/dx = - 3 cos (log x) 1/x - 4 sin (log x) * 1/x`
Multiplying both sides by x,
`x^2 (d^2 y)/dx^2 + x dy/dx` = −[3 cos (log x) + 4 sin (log x)]
`x^2 (d^2 y)/dx^2 + x dy/dx` = −y ...[From equation (1)]
`=> x^2 (d^2 y)/dx^2 + x dy/dx + y = 0`
Or, x2y2 + xy1 + y = 0
APPEARS IN
संबंधित प्रश्न
Find the second order derivative of the function.
x2 + 3x + 2
Find the second order derivative of the function.
x20
Find the second order derivative of the function.
e6x cos 3x
Find the second order derivative of the function.
tan–1 x
Find the second order derivative of the function.
log (log x)
Find the second order derivative of the function.
sin (log x)
If y = 5 cos x – 3 sin x, prove that `(d^2y)/(dx^2) + y = 0`.
If y = cos–1 x, find `(d^2y)/dx^2` in terms of y alone.
If y = 500e7x + 600e–7x, show that `(d^2y)/(dx^2)` = 49y.
If ey (x + 1) = 1, show that `(d^2y)/(dx^2) = (dy/dx)^2`.
Find `("d"^2"y")/"dx"^2`, if y = `"x"^-7`
Find `("d"^2"y")/"dx"^2`, if y = `"e"^"x"`
Find `("d"^2"y")/"dx"^2`, if y = log (x).
If ax2 + 2hxy + by2 = 0, then show that `("d"^2"y")/"dx"^2` = 0
tan–1(x2 + y2) = a
(x2 + y2)2 = xy
If x sin (a + y) + sin a cos (a + y) = 0, prove that `"dy"/"dx" = (sin^2("a" + y))/sin"a"`
If y = tan–1x, find `("d"^2y)/("dx"^2)` in terms of y alone.
If x2 + y2 + sin y = 4, then the value of `(d^2y)/(dx^2)` at the point (–2, 0) is ______.
Let for i = 1, 2, 3, pi(x) be a polynomial of degree 2 in x, p'i(x) and p''i(x) be the first and second order derivatives of pi(x) respectively. Let,
A(x) = `[(p_1(x), p_1^'(x), p_1^('')(x)),(p_2(x), p_2^'(x), p_2^('')(x)),(p_3(x), p_3^'(x), p_3^('')(x))]`
and B(x) = [A(x)]T A(x). Then determinant of B(x) ______
If x = A cos 4t + B sin 4t, then `(d^2x)/(dt^2)` is equal to ______.
`"Find" (d^2y)/(dx^2) "if" y=e^((2x+1))`
Find `(d^2y)/dx^2 if, y = e^((2x + 1))`
Find `(d^2y)/dx^2` if, `y = e^((2x + 1))`
Find `(d^2y)/dx^2` if, y = `e^((2x + 1))`
Find `(d^2y)/dx^2` if, `y = e^((2x + 1))`
Find `(d^2y)/dx^2, "if" y = e^((2x+1))`
Find `(d^2y)/dx^2` if, `y = e^((2x+1))`
Find `(d^2y)/(dx^2) "if", y = e^((2x + 1))`
Which expression defines the second order derivative of \[y\] with respect to \[x\]?
For \[y=\mathrm{A}\sin x+\mathrm{B}\cos x\], what is \[\frac{d^2y}{dx^2}\]?
After differentiating \[\sqrt{1-x^2}\cdot\frac{dy}{dx}=1\], which equation results?
What is \[\frac{d}{dx}\left(\sqrt{1-x^2}\right)\] in the differentiation for \[y=\sin^{-1}x\]?
What equation is obtained after multiplying through by \[\sqrt{1-x^2}\] in the derivation for \[y=\sin^{-1}x\]?
Differentiating \[(1-x^2)y_{1}^{2}=1\] gives which expression?
