Advertisements
Advertisements
प्रश्न
In the following example, verify that the given function is a solution of the corresponding differential equation.
| Solution | D.E. |
| xy = log y + k | y' (1 - xy) = y2 |
उत्तर
xy = log y + k
Differentiating w.r.t. x, we get
`x dy/dx+ y (1) = 1/y.dy/dx`
∴ `xy dy/dx+ y ^2 = dy/dx`
∴ `dy/dx- x y dy/dx = y^2`
∴ `(1-xy)dy/dx = y^2`
∴ `y' (1-xy) = y^2`
∴ Given function is a solution of the given differential equation.
APPEARS IN
संबंधित प्रश्न
Show that Ax2 + By2 = 1 is a solution of the differential equation x \[\left\{ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right\} = y\frac{dy}{dx}\]
Show that y = ax3 + bx2 + c is a solution of the differential equation \[\frac{d^3 y}{d x^3} = 6a\].
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 tangent at any point (x, y) of a curve makes an angle tan−1(2x + 3y) with x-axis. Find the equation of the curve if it passes through (1, 2).
The normal to a given curve at each point (x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4), find its equation.
Show that y = ae2x + be−x is a solution of the differential equation \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} - 2y = 0\]
Find the equation of the plane passing through the point (1, -2, 1) and perpendicular to the line joining the points A(3, 2, 1) and B(1, 4, 2).
Solve the following differential equation.
`dy/dx = x^2 y + y`
Solve the following differential equation.
`xy dy/dx = x^2 + 2y^2`
A solution of a differential equation which can be obtained from the general solution by giving particular values to the arbitrary constants is called ___________ solution.
Solve the differential equation (x2 – yx2)dy + (y2 + xy2)dx = 0
Solve: `("d"y)/("d"x) + 2/xy` = x2
`d/(dx)(tan^-1 (sqrt(1 + x^2) - 1)/x)` is equal to:
