Advertisements
Advertisements
प्रश्न
Solve the following differential equation.
`y^3 - dy/dx = x dy/dx`
Advertisements
उत्तर
`y^3 - dy/dx = x dy/dx`
∴ `y^3 = (1+x) dy/dx`
∴ `dx/((1+x)) = dy/y^3`
Integrating on both sides, we get
`intdx/(1+x )= int dy/y^3`
∴ `log | 1+x| = -1/(2y^2 )+c`
∴ 2y2 log | 1 + x | = – 1 + 2y2c
APPEARS IN
संबंधित प्रश्न
Show that y = ex (A cos x + B sin x) is the solution of the differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + 2y = 0\]
Verify that y2 = 4a (x + a) is a solution of the differential equations
\[y\left\{ 1 - \left( \frac{dy}{dx} \right)^2 \right\} = 2x\frac{dy}{dx}\]
Differential equation \[x\frac{dy}{dx} = 1, y\left( 1 \right) = 0\]
Function y = log x
x cos y dy = (xex log x + ex) dx
Solve the following differential equation:
\[xy\frac{dy}{dx} = 1 + x + y + xy\]
Find the solution of the differential equation cos y dy + cos x sin y dx = 0 given that y = \[\frac{\pi}{2}\], when x = \[\frac{\pi}{2}\]
In a bank principal increases at the rate of r% per year. Find the value of r if ₹100 double itself in 10 years (loge 2 = 0.6931).
x2 dy + y (x + y) dx = 0
Solve the following initial value problem:-
\[\frac{dy}{dx} - 3y \cot x = \sin 2x; y = 2\text{ when }x = \frac{\pi}{2}\]
Show that all curves for which the slope at any point (x, y) on it is \[\frac{x^2 + y^2}{2xy}\] are rectangular hyperbola.
Find the equation of the curve that passes through the point (0, a) and is such that at any point (x, y) on it, the product of its slope and the ordinate is equal to the abscissa.
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
\[\frac{dy}{dx} + Py = Q y^n , n > 2\] can be reduced to linear form by substituting
Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y \sin x = 1\], is
Solve the following differential equation.
`(dθ)/dt = − k (θ − θ_0)`
Solve the following differential equation.
xdx + 2y dx = 0
Solve the following differential equation.
y2 dx + (xy + x2 ) dy = 0
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
`d/(dx)(tan^-1 (sqrt(1 + x^2) - 1)/x)` is equal to:
Solve the differential equation
`x + y dy/dx` = x2 + y2
