Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
We have,
\[x\frac{dy}{dx} = x + y\]
\[ \Rightarrow \frac{dy}{dx} = \frac{x + y}{x}\]
This is a homogeneous differential equation .
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{x + vx}{x}\]
\[ \Rightarrow v + x\frac{dv}{dx} = 1 + v\]
\[ \Rightarrow x\frac{dv}{dx} = 1 + v - v\]
\[ \Rightarrow x\frac{dv}{dx} = 1\]
\[ \Rightarrow dv = \frac{1}{x}dx\]
Integrating both sides, we get
\[\int dv = \int\frac{1}{x}dx\]
\[ \Rightarrow v = \log \left| x \right| + C\]
\[\text{Putting }v = \frac{y}{x},\text{ we get}\]
\[ \Rightarrow \frac{y}{x} = \log \left| x \right| + C\]
\[ \Rightarrow y = x\log \left| x \right| + Cx \]
\[\text{ Hence, }y = x\log \left| x \right| + Cx\text{ is the required solution }.\]
APPEARS IN
संबंधित प्रश्न
If 1, `omega` and `omega^2` are the cube roots of unity, prove `(a + b omega + c omega^2)/(c + s omega + b omega^2) = omega^2`
Show that y = AeBx is a solution of the differential equation
Verify that \[y = e^{m \cos^{- 1} x}\] satisfies the differential equation \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} - m^2 y = 0\]
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x\frac{dy}{dx} = y\]
|
y = ax |
Differential equation \[\frac{d^2 y}{d x^2} - 3\frac{dy}{dx} + 2y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 3\] Function y = ex + e2x
Solve the following differential equation:
\[y e^\frac{x}{y} dx = \left( x e^\frac{x}{y} + y^2 \right)dy, y \neq 0\]
x2 dy + y (x + y) dx = 0
(x + 2y) dx − (2x − y) dy = 0
Solve the following initial value problem:-
\[\tan x\left( \frac{dy}{dx} \right) = 2x\tan x + x^2 - y; \tan x \neq 0\] given that y = 0 when \[x = \frac{\pi}{2}\]
Experiments show that radium disintegrates at a rate proportional to the amount of radium present at the moment. Its half-life is 1590 years. What percentage will disappear in one year?
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 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).
A curve is such that the length of the perpendicular from the origin on the tangent at any point P of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is \[y^2 - 2xy\frac{dy}{dx} - x^2 = 0\], and hence find the curve.
Write the differential equation obtained eliminating the arbitrary constant C in the equation xy = C2.
The differential equation satisfied by ax2 + by2 = 1 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 \[\frac{du}{dx} + P\left( x \right) u = Q\left( x \right)\]
Which of the following differential equations has y = C1 ex + C2 e−x as the general solution?
Solve the differential equation:
`"x"("dy")/("dx")+"y"=3"x"^2-2`
In the following example, verify that the given function is a solution of the corresponding differential equation.
| Solution | D.E. |
| y = xn | `x^2(d^2y)/dx^2 - n xx (xdy)/dx + ny =0` |
For the following differential equation find the particular solution.
`(x + 1) dy/dx − 1 = 2e^(−y)`,
when y = 0, x = 1
For the following differential equation find the particular solution.
`dy/ dx = (4x + y + 1),
when y = 1, x = 0
y dx – x dy + log x dx = 0
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 sec2y tan x dy + sec2x tan y dx = 0
Choose the correct alternative:
General solution of `y - x ("d"y)/("d"x)` = 0 is
The function y = cx is the solution of differential equation `("d"y)/("d"x) = y/x`
Solve the following differential equation
`y log y ("d"x)/("d"y) + x` = log y
Solve `x^2 "dy"/"dx" - xy = 1 + cos(y/x)`, x ≠ 0 and x = 1, y = `pi/2`
Solve the differential equation `"dy"/"dx" + 2xy` = y
The differential equation (1 + y2)x dx – (1 + x2)y dy = 0 represents a family of:
