Solve the differential equation xdx + 2ydy = 0

प्रश्न

बेरीज

उत्तर

xdx + 2ydy = 0

Integrating on both sides, we get

`int x "d"x + 2inty "d"y` = 0

∴ `x^2/2 + 2(y^2/2)^2` = c₁

∴ x² + 2y² = c, where c = 2c₁

shaalaa.com

Basic Concepts of Differential Equations

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

Q 4 Q 3 Q 5

पाठ 2.6: Differential Equations - Attempt the following questions II

APPEARS IN

एससीईआरटी महाराष्ट्र Mathematics and Statistics (Arts and Science) [English] Standard 12 Maharashtra State Board

पाठ 2.6 Differential Equations
Attempt the following questions II | Q 4

एससीईआरटी महाराष्ट्र Mathematics and Statistics (Commerce) [English] Standard 12 Maharashtra State Board

पाठ 1.8 Differential Equation and Applications
Q.4 | Q 4

संबंधित प्रश्‍न

\[\sqrt{1 + \left( \frac{dy}{dx} \right)^2} = \left( c\frac{d^2 y}{d x^2} \right)^{1/3}\]

\[\frac{d^4 y}{d x^4} = \left\{ c + \left( \frac{dy}{dx} \right)^2 \right\}^{3/2}\]

Form the differential equation representing the family of ellipses having centre at the origin and foci on x-axis.

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 \[\frac{d^3 y}{d x^3} = 6a\].

Verify that \[y = ce^{tan^{- 1}} x\] is a solution of the differential equation \[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + \left( 2x - 1 \right)\frac{dy}{dx} = 0\]

Show that the differential equation of which \[y = 2\left( x^2 - 1 \right) + c e^{- x^2}\] is a solution is \[\frac{dy}{dx} + 2xy = 4 x^3\]

For the following differential equation verify that the accompanying function is a solution:

Differential equation	Function
\[x^3 \frac{d^2 y}{d x^2} = 1\]	\[y = ax + b + \frac{1}{2x}\]

\[\frac{dy}{dx} + 2x = e^{3x}\]

\[\sin^4 x\frac{dy}{dx} = \cos x\]

\[\sqrt{1 - x^4} dy = x\ dx\]

\[\left( x^3 + x^2 + x + 1 \right)\frac{dy}{dx} = 2 x^2 + x\]

\[\sin\left( \frac{dy}{dx} \right) = k ; y\left( 0 \right) = 1\]

\[x\frac{dy}{dx} + 1 = 0 ; y \left( - 1 \right) = 0\]

xy (y + 1) dy = (x² + 1) dx

\[x\frac{dy}{dx} + y = y^2\]

x cos² y dx = y cos² x dy

\[\frac{dy}{dx} = \frac{x e^x \log x + e^x}{x \cos y}\]

\[y\sqrt{1 + x^2} + x\sqrt{1 + y^2}\frac{dy}{dx} = 0\]

tan y \[\frac{dy}{dx}\] = sin (x + y) + sin (x − y)

\[\frac{dy}{dx} + \frac{\cos x \sin y}{\cos y} = 0\]

y (1 + e^x) dy = (y + 1) e^x dx

\[\frac{dy}{dx} = 1 - x + y - xy\]

(y² + 1) dx − (x² + 1) dy = 0

dy + (x + 1) (y + 1) dx = 0

\[\left( x - 1 \right)\frac{dy}{dx} = 2 x^3 y\]

Solve the following differential equation:
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]

\[\frac{dr}{dt} = - rt, r\left( 0 \right) = r_0\]

\[\frac{dy}{dx} = 2xy, y\left( 0 \right) = 1\]

Find the particular solution of e^dy^/dx = x + 1, given that y = 3, when x = 0.

In a bank principal increases at the rate of r% per year. Find the value of r if ₹100 double itself in 10 years (log_e 2 = 0.6931).

\[\frac{dy}{dx} = \left( x + y \right)^2\]

\[\cos^2 \left( x - 2y \right) = 1 - 2\frac{dy}{dx}\]

\[\left( x + y + 1 \right)\frac{dy}{dx} = 1\]

\[\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}\]

3x² dy = (3xy + y²) dx

(x + 2y) dx − (2x − y) dy = 0

Solve the following initial value problem:
\[\frac{dy}{dx} + y \cot x = 4x\text{ cosec }x, y\left( \frac{\pi}{2} \right) = 0\]

Solve the following initial value problem:-

\[\frac{dy}{dx} - 3y \cot x = \sin 2x; y = 2\text{ when }x = \frac{\pi}{2}\]

Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation
\[y - x\frac{dy}{dx} = y^2 + \frac{dy}{dx}\]

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).

Find the equation of the curve which passes through the origin and has the slope x + 3y− 1 at any point (x, y) on it.

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 equation of the curve whose slope is given by \[\frac{dy}{dx} = \frac{2y}{x}; x > 0, y > 0\] and which passes through the point (1, 1) is

The differential equation
\[\frac{dy}{dx} + Py = Q y^n , n > 2\] can be reduced to linear form by substituting

Which of the following differential equations has y = C₁ e^x + C₂ e⁻^x as the general solution?

If x^myⁿ = (x + y)^m+n, prove that \[\frac{dy}{dx} = \frac{y}{x} .\]

Solve the differential equation:

`"x"("dy")/("dx")+"y"=3"x"^2-2`

Choose the correct option from the given alternatives:

The solution of `1/"x" * "dy"/"dx" = tan^-1 "x"` is

For the following differential equation find the particular solution.

`(x + 1) dy/dx − 1 = 2e^(−y)`,

when y = 0, x = 1