Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
We have,
\[\sqrt{a + x}dy + x\ dx = 0\]
\[ \Rightarrow \sqrt{a + x}dy = - xdx\]
\[ \Rightarrow dy = \frac{- x}{\sqrt{a + x}}dx\]
\[ \Rightarrow dy = - \frac{\left( x + a - a \right)}{\sqrt{a + x}}dx\]
\[ \Rightarrow dy = - \left( \sqrt{a + x} - \frac{a}{\sqrt{a + x}} \right)dx\]
Integrating both sides, we get
\[\int dy = - \int\left( \sqrt{a + x} - \frac{a}{\sqrt{a + x}} \right)dx\]
\[ \Rightarrow y = - \frac{2 \left( a + x \right)^\frac{3}{2}}{3} + 2a\sqrt{a + x} + C\]
\[ \Rightarrow y + \frac{2}{3} \left( a + x \right)^\frac{3}{2} - 2a\sqrt{a + x} = C\]
\[\text{ Hence, }y + \frac{2}{3} \left( a + x \right)^\frac{3}{2} - 2a\sqrt{a + x} = \text{C is the solution to the given differential equation.}\]
APPEARS IN
संबंधित प्रश्न
Show that the differential equation of which y = 2(x2 − 1) + \[c e^{- x^2}\] is a solution, is \[\frac{dy}{dx} + 2xy = 4 x^3\]
Form the differential equation representing the family of ellipses having centre at the origin and foci on x-axis.
Show that the function y = A cos 2x − B sin 2x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + 4y = 0\].
Show that y = ax3 + bx2 + c is a solution of the differential equation \[\frac{d^3 y}{d x^3} = 6a\].
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x^3 \frac{d^2 y}{d x^2} = 1\]
|
\[y = ax + b + \frac{1}{2x}\]
|
Differential equation \[\frac{dy}{dx} + y = 2, y \left( 0 \right) = 3\] Function y = e−x + 2
x cos2 y dx = y cos2 x dy
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 + y) (dx − dy) = dx + dy
2xy dx + (x2 + 2y2) dy = 0
The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present.
Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation
\[y - x\frac{dy}{dx} = y^2 + \frac{dy}{dx}\]
Find the curve for which the intercept cut-off by a tangent on x-axis is equal to four times the ordinate of the point of contact.
Show that the equation of the curve whose slope at any point is equal to y + 2x and which passes through the origin is y + 2 (x + 1) = 2e2x.
The rate of increase of bacteria in a culture is proportional to the number of bacteria present and it is found that the number doubles in 6 hours. Prove that the bacteria becomes 8 times at the end of 18 hours.
The integrating factor of the differential equation (x log x)
\[\frac{dy}{dx} + y = 2 \log x\], is given by
The differential equation of the ellipse \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = C\] is
Which of the following differential equations has y = C1 ex + C2 e−x as the general solution?
Form the differential equation of the family of circles having centre on y-axis and radius 3 unit.
If a + ib = `("x" + "iy")/("x" - "iy"),` prove that `"a"^2 +"b"^2 = 1` and `"b"/"a" = (2"xy")/("x"^2 - "y"^2)`
For the following differential equation find the particular solution.
`(x + 1) dy/dx − 1 = 2e^(−y)`,
when y = 0, x = 1
Solve the following differential equation.
`x^2 dy/dx = x^2 +xy - y^2`
The solution of `dy/ dx` = 1 is ______.
Solve
`dy/dx + 2/ x y = x^2`
Select and write the correct alternative from the given option for the question
The differential equation of y = Ae5x + Be–5x is
Solve the differential equation `("d"y)/("d"x) + y` = e−x
Solve the following differential equation `("d"y)/("d"x)` = x2y + y
Solve the following differential equation y log y = `(log y - x) ("d"y)/("d"x)`
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
Solve `x^2 "dy"/"dx" - xy = 1 + cos(y/x)`, x ≠ 0 and x = 1, y = `pi/2`
