Advertisements
Advertisements
प्रश्न
Solve:
(x + y) dy = a2 dx
Advertisements
उत्तर
(x + y) dy = a2 dx
∴ `dy/dx = a^2/(x+y)` ...(i)
Put x + y = t ...(ii)
∴ y = t - x
Differentiating w.r.t. x, we get
∴ `dy/dx = dt /dx -1` ....(iii)
Substituting (ii) and (iii) in (i), we get
`dt/dx -1 = a^2/t`
∴ `dt/dx = a^2/t + 1`
∴ `dt/dx = (a^2+t)/t`
∴ `t/(a^2+t) dt = dx`
Integrating on both sides, we get
`int ((a^2+t) - a^2)/(a^2+ t) dt = int dx`
∴ `int 1 dt- a^2int 1/(a^2+t) dt = int dx`
∴ t - a2 log |a2 + t| = x + c1
∴ x + y - a2 log |a2 + x + y| = x + c1
∴ y - a2 log |a2 + x + y| = c1
∴ y - c1 = a2 log |a2 + x + y|
∴ `y/a^2 - c_1/a^2 = log |a^2 + x + y|`
∴ `a^2 + x + y = e^(a^(y/2). e^(a^((-c1)/2)`
∴ `a^2 + x + y = ce^(a^(y/2) ` … `[ c =e^(a^((-c1)/2)]]`
APPEARS IN
संबंधित प्रश्न
Show that the differential equation of which y = 2(x2 − 1) + \[c e^{- x^2}\] is a solution, is \[\frac{dy}{dx} + 2xy = 4 x^3\]
Verify that \[y = ce^{tan^{- 1}} x\] is a solution of the differential equation \[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + \left( 2x - 1 \right)\frac{dy}{dx} = 0\]
Differential equation \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} = 0, y \left( 0 \right) = 2, y'\left( 0 \right) = 1\]
Function y = ex + 1
Differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 2\] Function y = xex + ex
(1 − x2) dy + xy dx = xy2 dx
Solve the following differential equation:
\[y\left( 1 - x^2 \right)\frac{dy}{dx} = x\left( 1 + y^2 \right)\]
Solve the following differential equation:
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]
Solve the following initial value problem:-
\[x\frac{dy}{dx} - y = \log x, y\left( 1 \right) = 0\]
In a simple circuit of resistance R, self inductance L and voltage E, the current `i` at any time `t` is given by L \[\frac{di}{dt}\]+ R i = E. If E is constant and initially no current passes through the circuit, prove that \[i = \frac{E}{R}\left\{ 1 - e^{- \left( R/L \right)t} \right\}.\]
The decay rate of radium at any time t is proportional to its mass at that time. Find the time when the mass will be halved of its initial mass.
The slope of the tangent at a point P (x, y) on a curve is \[\frac{- x}{y}\]. If the curve passes through the point (3, −4), find the equation of the curve.
Find the equation of the curve which passes through the point (1, 2) and the distance between the foot of the ordinate of the point of contact and the point of intersection of the tangent with x-axis is twice the abscissa of the point of contact.
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 all curves for which the slope at any point (x, y) on it is \[\frac{x^2 + y^2}{2xy}\] are rectangular hyperbola.
The slope of a curve at each of its points is equal to the square of the abscissa of the point. Find the particular curve through the point (−1, 1).
The differential equation of the ellipse \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = C\] is
The differential equation satisfied by ax2 + by2 = 1 is
Solve the following differential equation : \[\left( \sqrt{1 + x^2 + y^2 + x^2 y^2} \right) dx + xy \ dy = 0\].
Show that y = ae2x + be−x is a solution of the differential equation \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} - 2y = 0\]
Solve the following differential equation.
`dy/dx = x^2 y + y`
Solve the following differential equation.
`(dθ)/dt = − k (θ − θ_0)`
Solve the following differential equation.
(x2 − y2 ) dx + 2xy dy = 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 `"dy"/"dx"` = 1 + x + y2 + xy2, when y = 0, x = 0.
Solve the differential equation
`x + y dy/dx` = x2 + y2
