Solve the following differential equation yxdydx = x2 + 2y2 - Mathematics and Statistics

Question

Solve the following differential equation

`yx ("d"y)/("d"x)` = x² + 2y²

Sum

Solution

`yx ("d"y)/("d"x)` = x² + 2y²

∴ `("d"y)/("d"x) = (x^2 + 2y^2)/(xy)` ......(i)

Put y = vx ......(ii)

Differentiating w.r.t. x, we get

`("d"y)/("d"x) = "v" + x "dv"/("d"x)` ......(iii)

Substituting (ii) and (iii) in (i), we get

`"v" + x "dv"/("d"x) = (x^2 + 2"v"^2x^2)/(x("v"x))`

∴ `"v" + x "dv"/("d"x) = (x^2(1 + 2"v"^2))/(x^2"v")`

∴ `x "dv"/("d"x) = (1 + 2"v"^2)/"v" - "v"`

= `(1 + "v"^2)/"v"`

∴ `"v"/(1 + "v"^2) "dv" = 1/x "d"x`

Integrating on both sides, we get

`1/ int (2"v")/(1 +"v"^2) "dv" = int "dv"/x`

∴ `1/2 log|1 + "v"^2|` = log |x| + log |c|

∴ log |1 + c²| = 2 og |x| + 2log |c|

= log |x²| + log |c²|

∴ log |1 + v²| = log |c²x²|

∴ 1 + v² = c²x²

∴ `1 + y^2/x^2` = c²x²

∴ x² + y² = c²x⁴

shaalaa.com

Differential Equations

Is there an error in this question or solution?

Q 6 Q 5 Q 7

Chapter 1.8: Differential Equation and Applications - Q.5

APPEARS IN

SCERT Maharashtra Mathematics and Statistics (Commerce) [English] 12 Standard HSC

Chapter 1.8 Differential Equation and Applications
Q.5 | Q 6

SCERT Maharashtra Mathematics and Statistics (Arts and Science) [English] 12 Standard HSC

Chapter 2.6 Differential Equations
Attempt the following questions III | Q 6

RELATED QUESTIONS

Prove that:

`int_0^(2a)f(x)dx = int_0^af(x)dx + int_0^af(2a - x)dx`

Solve the equation for x: `sin^(-1) 5/x + sin^(-1) 12/x = pi/2, x != 0`

\[\frac{d^3 x}{d t^3} + \frac{d^2 x}{d t^2} + \left( \frac{dx}{dt} \right)^2 = e^t\]

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

\[\sqrt[3]{\frac{d^2 y}{d x^2}} = \sqrt{\frac{dy}{dx}}\]

\[x^2 \left( \frac{d^2 y}{d x^2} \right)^3 + y \left( \frac{dy}{dx} \right)^4 + y^4 = 0\]

Find the differential equation of all the parabolas with latus rectum '4a' and whose axes are parallel to x-axis.

Verify that y² = 4ax is a solution of the differential equation y = x \[\frac{dy}{dx} + a\frac{dx}{dy}\]

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 \[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + \left( 2x - 1 \right)\frac{dy}{dx} = 0\]

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

Differential equation	Function
\[y = \left( \frac{dy}{dx} \right)^2\]	\[y = \frac{1}{4} \left( x \pm a \right)^2\]

Differential equation \[x\frac{dy}{dx} = 1, y\left( 1 \right) = 0\]

Function y = log x

Differential equation \[\frac{dy}{dx} = y, y\left( 0 \right) = 1\]
Function y = e^x

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 = e^x + 1

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

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

\[\frac{dy}{dx} = \log x\]

\[\sqrt{a + x} dy + x\ dx = 0\]

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

\[5\frac{dy}{dx} = e^x y^4\]

Solve the differential equation \[\frac{dy}{dx} = e^{x + y} + x^2 e^y\].

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

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

\[\frac{dy}{dx} = \frac{x\left( 2 \log x + 1 \right)}{\sin y + y \cos y}\]

(1 − x²) dy + xy dx = xy² dx

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

(y² + 1) dx − (x² + 1) dy = 0

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

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

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

Solve the following differential equation:
\[y e^\frac{x}{y} dx = \left( x e^\frac{x}{y} + y^2 \right)dy, y \neq 0\]

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

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

The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after `t` seconds.

If y(x) is a solution of the different equation \[\left( \frac{2 + \sin x}{1 + y} \right)\frac{dy}{dx} = - \cos x\] and y(0) = 1, then find the value of y(π/2).

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