English

Solve: (x + y) dy = a2 dx

Advertisements
Advertisements

Question

Solve:

(x + y) dy = a2 dx

Sum
Advertisements

Solution

(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)]]`

shaalaa.com
  Is there an error in this question or solution?
Chapter 8: Differential Equation and Applications - Miscellaneous Exercise 8 [Page 173]

APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
Chapter 8 Differential Equation and Applications
Miscellaneous Exercise 8 | Q 4.07 | Page 173

RELATED QUESTIONS

\[y\frac{d^2 x}{d y^2} = y^2 + 1\]

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


Show that the function y = A cos x + B sin x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + y = 0\]


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

Differential equation Function
\[x + y\frac{dy}{dx} = 0\]
\[y = \pm \sqrt{a^2 - x^2}\]

\[\frac{dy}{dx} = \tan^{- 1} x\]


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

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

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

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

(1 + x2) dy = xy dx


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

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

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

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

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

3x2 dy = (3xy + y2) dx


Solve the following differential equations:
\[\frac{dy}{dx} = \frac{y}{x}\left\{ \log y - \log x + 1 \right\}\]


Find the particular solution of the differential equation \[\frac{dy}{dx} = \frac{xy}{x^2 + y^2}\] given that y = 1 when x = 0.

 


The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present.


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.


Find the equation of the curve such that the portion of the x-axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2).


The solution of the differential equation \[\frac{dy}{dx} - \frac{y\left( x + 1 \right)}{x} = 0\] is given by


The differential equation satisfied by ax2 + by2 = 1 is


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


Determine the order and degree of the following differential equations.

Solution D.E.
ax2 + by2 = 5 `xy(d^2y)/dx^2+ x(dy/dx)^2 = y dy/dx`

Solve the following differential equation.

`dy/dx + 2xy = x`


y2 dx + (xy + x2)dy = 0


Solve the following differential equation y2dx + (xy + x2) dy = 0


lf the straight lines `ax + by + p` = 0 and `x cos alpha + y sin alpha = p` are inclined at an angle π/4 and concurrent with the straight line `x sin alpha - y cos alpha` = 0, then the value of `a^2 + b^2` is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×