Advertisements
Advertisements
प्रश्न
In the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:-
`y=sqrt(a^2-x^2)` `x+y(dy/dx)=0`
Advertisements
उत्तर
We have,
`x+y(dy)/(dx)=0 .............(1)`
Now,
`y=sqrt(a^2-x^2)`
`rArry'=(-x)/(sqrt(a^2-x^2))`
Putting the above value in (1), we get
`"LHS" =x+y((-x)/(sqrt(a^2-x^2)))`
`=x+sqrt(a^2-x^2)xx(-x)/(sqrt(a^2-x^2))`
`=x-x=0=" RHS"`
Thus, `y=sqrt(a^2-x^2)` is the solution of the given differential equation.
APPEARS IN
संबंधित प्रश्न
Hence, the given function is the solution to the given differential equation. \[\frac{c - x}{1 + cx}\] is a solution of the differential equation \[(1+x^2)\frac{dy}{dx}+(1+y^2)=0\].
Show that y = e−x + ax + b is solution of the differential equation\[e^x \frac{d^2 y}{d x^2} = 1\]
Differential equation \[x\frac{dy}{dx} = 1, y\left( 1 \right) = 0\]
Function y = log x
Solve the following differential equation:
\[xy\frac{dy}{dx} = 1 + x + y + xy\]
Solve the following differential equation:
\[y e^\frac{x}{y} dx = \left( x e^\frac{x}{y} + y^2 \right)dy, y \neq 0\]
\[x^2 \frac{dy}{dx} = x^2 + xy + y^2 \]
(x + 2y) dx − (2x − y) dy = 0
A population grows at the rate of 5% per year. How long does it take for the population to double?
Find the equation of the curve such that the portion of the x-axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2).
Write the differential equation obtained eliminating the arbitrary constant C in the equation xy = C2.
Find the solution of the differential equation
\[x\sqrt{1 + y^2}dx + y\sqrt{1 + x^2}dy = 0\]
The solution of the differential equation y1 y3 = y22 is
The differential equation of the ellipse \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = C\] is
Which of the following transformations reduce the differential equation \[\frac{dz}{dx} + \frac{z}{x}\log z = \frac{z}{x^2} \left( \log z \right)^2\] into the form \[\frac{du}{dx} + P\left( x \right) u = Q\left( x \right)\]
What is integrating factor of \[\frac{dy}{dx}\] + y sec x = tan x?
For each of the following differential equations find the particular solution.
(x − y2 x) dx − (y + x2 y) dy = 0, when x = 2, y = 0
The integrating factor of the differential equation `dy/dx - y = x` is e−x.
State whether the following is True or False:
The degree of a differential equation is the power of the highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any.
Solve the differential equation:
dr = a r dθ − θ dr
Solve:
(x + y) dy = a2 dx
y2 dx + (xy + x2)dy = 0
Solve the differential equation (x2 – yx2)dy + (y2 + xy2)dx = 0
Choose the correct alternative:
Solution of the equation `x("d"y)/("d"x)` = y log y is
Verify y = `a + b/x` is solution of `x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0
y = `a + b/x`
`(dy)/(dx) = square`
`(d^2y)/(dx^2) = square`
Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`
= `x square + 2 square`
= `square`
Hence y = `a + b/x` is solution of `square`
Solve the following differential equation `("d"y)/("d"x)` = cos(x + y)
Solution: `("d"y)/("d"x)` = cos(x + y) ......(1)
Put `square`
∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`
∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`
∴ (1) becomes `"dv"/("d"x) - 1` = cos v
∴ `"dv"/("d"x)` = 1 + cos v
∴ `square` dv = dx
Integrating, we get
`int 1/(1 + cos "v") "d"v = int "d"x`
∴ `int 1/(2cos^2 ("v"/2)) "dv" = int "d"x`
∴ `1/2 int square "dv" = int "d"x`
∴ `1/2* (tan("v"/2))/(1/2)` = x + c
∴ `square` = x + c
Given that `"dy"/"dx"` = yex and x = 0, y = e. Find the value of y when x = 1.
Given that `"dy"/"dx" = "e"^-2x` and y = 0 when x = 5. Find the value of x when y = 3.
The differential equation (1 + y2)x dx – (1 + x2)y dy = 0 represents a family of:
