मराठी

X D Y D X = X + Y

Advertisements
Advertisements

प्रश्न

\[x\frac{dy}{dx} = x + y\]
बेरीज
Advertisements

उत्तर

We have,
\[x\frac{dy}{dx} = x + y\]
\[ \Rightarrow \frac{dy}{dx} = \frac{x + y}{x}\]
This is a homogeneous differential equation . 
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{x + vx}{x}\]
\[ \Rightarrow v + x\frac{dv}{dx} = 1 + v\]
\[ \Rightarrow x\frac{dv}{dx} = 1 + v - v\]
\[ \Rightarrow x\frac{dv}{dx} = 1\]
\[ \Rightarrow dv = \frac{1}{x}dx\]
Integrating both sides, we get 
\[\int dv = \int\frac{1}{x}dx\]
\[ \Rightarrow v = \log \left| x \right| + C\]
\[\text{Putting }v = \frac{y}{x},\text{ we get}\]
\[ \Rightarrow \frac{y}{x} = \log \left| x \right| + C\]
\[ \Rightarrow y = x\log \left| x \right| + Cx \]
\[\text{ Hence, }y = x\log \left| x \right| + Cx\text{ is the required solution }.\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 21: Differential Equations - Exercise 22.09 [पृष्ठ ८३]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 21 Differential Equations
Exercise 22.09 | Q 4 | पृष्ठ ८३

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

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


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

 


Hence, the given function is the solution to the given differential equation. \[\frac{c - x}{1 + cx}\] is a solution of the differential equation \[(1+x^2)\frac{dy}{dx}+(1+y^2)=0\].


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


\[\frac{1}{x}\frac{dy}{dx} = \tan^{- 1} x, x \neq 0\]

\[\frac{dy}{dx} = \cos^3 x \sin^2 x + x\sqrt{2x + 1}\]

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

x cos y dy = (xex log x + ex) dx


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

Solve the following differential equation: 
(xy2 + 2x) dx + (x2 y + 2y) dy = 0


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


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

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

In a culture the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present.


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

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

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


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

Solve the following initial value problem:-
\[\tan x\left( \frac{dy}{dx} \right) = 2x\tan x + x^2 - y; \tan x \neq 0\] given that y = 0 when \[x = \frac{\pi}{2}\]


A bank pays interest by continuous compounding, that is, by treating the interest rate as the instantaneous rate of change of principal. Suppose in an account interest accrues at 8% per year, compounded continuously. Calculate the percentage increase in such an account over one year.


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 passing through the point \[\left( 1, \frac{\pi}{4} \right)\]  and tangent at any point of which makes an angle tan−1  \[\left( \frac{y}{x} - \cos^2 \frac{y}{x} \right)\] with x-axis.


At every point on a curve the slope is the sum of the abscissa and the product of the ordinate and the abscissa, and the curve passes through (0, 1). Find the equation of the curve.


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


Which of the following differential equations has y = C1 ex + C2 ex as the general solution?


Solve the following differential equation.

y2 dx + (xy + x2 ) dy = 0


Solve the following differential equation.

y dx + (x - y2 ) dy = 0


Choose the correct alternative.

The differential equation of y = `k_1 + k_2/x` is


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


The integrating factor of the differential equation `dy/dx - y = x` is e−x.


Solve the differential equation:

`e^(dy/dx) = x`


Solve `("d"y)/("d"x) = (x + y + 1)/(x + y - 1)` when x = `2/3`, y = `1/3`


Solve the following differential equation

`yx ("d"y)/("d"x)` = x2 + 2y2 


A solution of differential equation which can be obtained from the general solution by giving particular values to the arbitrary constant is called ______ solution


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


If `y = log_2 log_2(x)` then `(dy)/(dx)` =


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×