English

D Y D X + 1 = E X + Y

Advertisements
Advertisements

Question

\[\frac{dy}{dx} + 1 = e^{x + y}\]
Sum
Advertisements

Solution

\[\frac{dy}{dx} + 1 = e^{x + y}\]                .....(1)
Let x + y = t
\[\Rightarrow 1 + \frac{dy}{dx} = \frac{dt}{dx}\]
Substituting the value of x + y = t and \[1 + \frac{dy}{dx} = \frac{dt}{dx}\] in (1), we get
\[\frac{dt}{dx} = e^t \]
\[ \Rightarrow e^{- t} dt = dx\]
\[ \Rightarrow - e^{- t} = x + C\]
\[ \Rightarrow - e^{- \left( x + y \right)} = x + C ...........\left[ \because t = x + y \right]\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 21: Differential Equations - Exercise 22.08 [Page 66]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 21 Differential Equations
Exercise 22.08 | Q 11 | Page 66

RELATED QUESTIONS

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


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


Verify that \[y = e^{m \cos^{- 1} x}\] satisfies the differential equation \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} - m^2 y = 0\]


Verify that y = log \[\left( x + \sqrt{x^2 + a^2} \right)^2\]  satisfies the differential equation \[\left( a^2 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = 0\]


(sin x + cos x) dy + (cos x − sin x) dx = 0


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

(1 + x2) dy = xy dx


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

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

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

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

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

tan y dx + sec2 y tan x dy = 0


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

Solve the following differential equation:
\[\text{ cosec }x \log y \frac{dy}{dx} + x^2 y^2 = 0\]


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

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

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


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

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

The rate of growth of a population is proportional to the number present. If the population of a city doubled in the past 25 years, and the present population is 100000, when will the city have a population of 500000?


Write the differential equation representing the family of straight lines y = Cx + 5, where C is an arbitrary constant.


Which of the following is the integrating factor of (x log x) \[\frac{dy}{dx} + y\] = 2 log x?


Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y \sin x = 1\], is


In the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:-

y = ex + 1            y'' − y' = 0


Form the differential equation of the family of circles having centre on y-axis and radius 3 unit.


Find the particular solution of the differential equation `"dy"/"dx" = "xy"/("x"^2+"y"^2),`given that y = 1 when x = 0


Solve the following differential equation.

y2 dx + (xy + x2 ) dy = 0


Solve the following differential equation.

`xy  dy/dx = x^2 + 2y^2`


Solve the following differential equation.

`dy/dx + 2xy = x`


Choose the correct alternative.

Bacteria increases at the rate proportional to the number present. If the original number M doubles in 3 hours, then the number of bacteria will be 4M in


Solve

`dy/dx + 2/ x y = x^2`


y dx – x dy + log x dx = 0


The function y = ex is solution  ______ of differential equation


Solve the following differential equation `("d"y)/("d"x)` = x2y + y


Integrating factor of the differential equation `"dy"/"dx" - y` = cos x is ex.


Solve the differential equation `"dy"/"dx"` = 1 + x + y2 + xy2, when y = 0, x = 0.


The value of `dy/dx` if y = |x – 1| + |x – 4| at x = 3 is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×