हिंदी

If (a+bx)eyx = x then prove that xd2ydx2=(aa+bx)2. - Mathematics

Advertisements
Advertisements

प्रश्न

If `(a + bx)e^(y/x)` = x then prove that `x(d^2y)/(dx^2) = (a/(a + bx))^2`.

योग

उत्तर

`y/x = log_e (x/(a + bx))` = loge x – loge (a + bx)

On differentiating with respect to x, we get

`\implies (x dy/dx - y)/x^2 = 1/x - 1/(a + bx) d/dx(a + bx) = 1/x - b/(a + bx)`

`\implies x dy/dx - y = x^2(1/x - b/(a + bx)) = (ax)/(a + bx)`

On differentiating again with respect to x, we get

`\implies x (d^2y)/(dx^2) + dy/dx - dy/dx = ((a + bx)a - ax(b))/(a + bx)^2`

`\implies x (d^2y)/(dx^2) = (a/(a + bx))^2`.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2023-2024 (March) Board Sample Paper

वीडियो ट्यूटोरियलVIEW ALL [3]

संबंधित प्रश्न

Write the degree of the differential equation `x^3((d^2y)/(dx^2))^2+x(dy/dx)^4=0`


Determine the order and degree (if defined) of the differential equation:

`((ds)/(dt))^4 + 3s  (d^2s)/(dt^2) = 0`


The order of the differential equation `2x^2 (d^2y)/(dx^2) - 3 (dy)/(dx) + y = 0` is ______.


\[s^2 \frac{d^2 t}{d s^2} + st\frac{dt}{ds} = s\]

(xy2 + x) dx + (y − x2y) dy = 0


\[\frac{d^2 y}{d x^2} = \left( \frac{dy}{dx} \right)^{2/3}\]

\[\frac{d^2 y}{d x^2} + 3 \left( \frac{dy}{dx} \right)^2 = x^2 \log\left( \frac{d^2 y}{d x^2} \right)\]

\[\frac{d^3 y}{d x^3} + \frac{d^2 y}{d x^2} + \frac{dy}{dx} + y \sin y = 0\]

Write the order of the differential equation of the family of circles touching X-axis at the origin.


What is the degree of the following differential equation?

\[5x \left( \frac{dy}{dx} \right)^2 - \frac{d^2 y}{d x^2} - 6y = \log x\]

Write the degree of the differential equation \[x^3 \left( \frac{d^2 y}{d x^2} \right)^2 + x \left( \frac{dy}{dx} \right)^4 = 0\]


The degree of the differential equation \[\left( \frac{d^2 y}{d x^2} \right)^3 + \left( \frac{dy}{dx} \right)^2 + \sin\left( \frac{dy}{dx} \right) + 1 = 0\], is


The order of the differential equation \[2 x^2 \frac{d^2 y}{d x^2} - 3\frac{dy}{dx} + y = 0\], is


Write the order and degree of the differential equation `((d^4"y")/(d"x"^4))^2 =  [ "x" + ((d"y")/(d"x"))^2]^3`.


Determine the order and degree of the following differential equation:

`("d"^2"y")/"dx"^2 + "dy"/"dx" + "x" = sqrt(1 + ("d"^3"y")/"dx"^3)`


Determine the order and degree of the following differential equation:

`("d"^4"y")/"dx"^4 + sin ("dy"/"dx") = 0`


Determine the order and degree of the following differential equations.

`(y''')^2 + 2(y'')^2 + 6y' + 7y = 0`


Determine the order and degree of the following differential equations.

`dy/dx = 7 (d^2y)/dx^2`


State whether the following is True or False:

The order of highest derivative occurring in the differential equation is called degree of the differential equation.


State whether the following is True or False:

The degree of the differential equation `e^((dy)/(dx)) = dy/dx +c` is not defined.


Order and degree of differential equation are always ______ integers


The power of highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any is called ______ of the differential equation


The order and degree of the differential equation `[1 + 1/("dy"/"dx")^2]^(5/3) = 5 ("d"^2y)/"dx"^2` are respectively.


The order and degree of the differential equation `(dy/dx)^3 + ((d^3y)/dx^3) + xy = 0` are respectively ______


The degree of the differential equation `(1 + "dy"/"dx")^3 = (("d"^2y)/("d"x^2))^2` is ______.


The order and degree of the differential equation `(("d"^3y)/("d"x^3))^2 - 3 ("d"^2y)/("d"x^2) + 2(("d"y)/("d"x))^4` = y4 are ______.


Determine the order and degree of the following differential equation:

`(d^2y)/(dx^2) + x((dy)/(dx)) + y` = 2 sin x


The degree and order of the differential equation `[1 + (dy/dx)^3]^(7/3) = 7((d^2y)/(dx^2))` respectively are ______.


The degree of the differential equation `[1 + (dy/dx)^2]^3 = ((d^2y)/(dx^2))^2` is ______.


Find the order and degree of the differential equation

`sqrt(1 + 1/(dy/dx)^2) = ((d^2y)/(dx^2))^(3/2)`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×