Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
We have,
\[\frac{dy}{dx} = x \log x\]
\[ \Rightarrow dy = \left( x \log x \right)dx\]
Integrating both sides, we get
\[\int dy = \int\left( x \log x \right)dx\]
\[ \Rightarrow y = \log x\int x\ dx - \int\left[ \frac{d}{dx}\left( \log x \right)\int x\ dx \right]dx\]
\[ \Rightarrow y = \log x \times \frac{x^2}{2} - \int\left( \frac{1}{x} \times \frac{x^2}{2} \right)dx\]
\[ \Rightarrow y = \frac{1}{2} x^2 \log x - \int\frac{x}{2}dx\]
\[ \Rightarrow y = \frac{1}{2} x^2 \log x - \frac{x^2}{4} + C\]
\[ \text{ Hence, }y = \frac{1}{2} x^2 \log x - \frac{x^2}{4} + \text{C is the solution to the given differential equation.}\]
APPEARS IN
संबंधित प्रश्न
Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.
Verify that y2 = 4ax is a solution of the differential equation y = x \[\frac{dy}{dx} + a\frac{dx}{dy}\]
Differential equation \[x\frac{dy}{dx} = 1, y\left( 1 \right) = 0\]
Function y = log x
Differential equation \[\frac{d^2 y}{d x^2} - 3\frac{dy}{dx} + 2y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 3\] Function y = ex + e2x
Solve the following differential equation:
\[y e^\frac{x}{y} dx = \left( x e^\frac{x}{y} + y^2 \right)dy, y \neq 0\]
Solve the differential equation \[\left( 1 + x^2 \right)\frac{dy}{dx} + \left( 1 + y^2 \right) = 0\], given that y = 1, when x = 0.
Find the particular solution of the differential equation \[\frac{dy}{dx} = - 4x y^2\] given that y = 1, when x = 0.
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).
3x2 dy = (3xy + y2) dx
Solve the following initial value problem:-
\[dy = \cos x\left( 2 - y\text{ cosec }x \right)dx\]
The x-intercept of the tangent line to a curve is equal to the ordinate of the point of contact. Find the particular curve through the point (1, 1).
Define a differential equation.
Write the differential equation representing the family of straight lines y = Cx + 5, where C is an arbitrary constant.
The differential equation obtained on eliminating A and B from y = A cos ωt + B sin ωt, is
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 solution of the differential equation \[\frac{dy}{dx} = \frac{ax + g}{by + f}\] represents a circle when
In the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:-
`y=sqrt(a^2-x^2)` `x+y(dy/dx)=0`
Determine the order and degree of the following differential equations.
| Solution | D.E. |
| y = 1 − logx | `x^2(d^2y)/dx^2 = 1` |
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` |
Find the differential equation whose general solution is
x3 + y3 = 35ax.
Solve the following differential equation.
`dy/dx + 2xy = x`
Choose the correct alternative.
The integrating factor of `dy/dx - y = e^x `is ex, then its solution is
Solve the differential equation:
`e^(dy/dx) = x`
Solve:
(x + y) dy = a2 dx
`dy/dx = log x`
y dx – x dy + log x dx = 0
Select and write the correct alternative from the given option for the question
Bacterial increases at the rate proportional to the number present. If original number M doubles in 3 hours, then number of bacteria will be 4M in
For the differential equation, find the particular solution (x – y2x) dx – (y + x2y) dy = 0 when x = 2, y = 0
For the differential equation, find the particular solution
`("d"y)/("d"x)` = (4x +y + 1), when y = 1, x = 0
Solve the following differential equation
sec2 x tan y dx + sec2 y tan x dy = 0
Solution: sec2 x tan y dx + sec2 y tan x dy = 0
∴ `(sec^2x)/tanx "d"x + square` = 0
Integrating, we get
`square + int (sec^2y)/tany "d"y` = log c
Each of these integral is of the type
`int ("f'"(x))/("f"(x)) "d"x` = log |f(x)| + log c
∴ the general solution is
`square + log |tan y|` = log c
∴ log |tan x . tan y| = log c
`square`
This is the general solution.
Solve the following differential equation `("d"y)/("d"x)` = cos(x + y)
Solution: `("d"y)/("d"x)` = cos(x + y) ......(1)
Put `square`
∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`
∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`
∴ (1) becomes `"dv"/("d"x) - 1` = cos v
∴ `"dv"/("d"x)` = 1 + cos v
∴ `square` dv = dx
Integrating, we get
`int 1/(1 + cos "v") "d"v = int "d"x`
∴ `int 1/(2cos^2 ("v"/2)) "dv" = int "d"x`
∴ `1/2 int square "dv" = int "d"x`
∴ `1/2* (tan("v"/2))/(1/2)` = x + c
∴ `square` = x + c
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
