Advertisements
Advertisements
Question
Advertisements
Solution
We have,
\[\frac{dy}{dx} = \left( e^x + 1 \right) y\]
\[ \Rightarrow \frac{1}{y}dy = \left( e^x + 1 \right) dx\]
Integrating both sides, we get
\[\int\frac{1}{y}dy = \int\left( e^x + 1 \right) dx\]
\[ \Rightarrow \log \left| y \right| = e^x + x + C\]
\[\text{ Hence, }\log \left| y \right| = e^x + x +\text{ C is the required solution .} \]
APPEARS IN
RELATED QUESTIONS
Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.
Verify that y = \[\frac{a}{x} + b\] is a solution of the differential equation
\[\frac{d^2 y}{d x^2} + \frac{2}{x}\left( \frac{dy}{dx} \right) = 0\]
Hence, the given function is the solution to the given differential equation. \[\frac{c - x}{1 + cx}\] is a solution of the differential equation \[(1+x^2)\frac{dy}{dx}+(1+y^2)=0\].
Show that the differential equation of which \[y = 2\left( x^2 - 1 \right) + c e^{- x^2}\] is a solution is \[\frac{dy}{dx} + 2xy = 4 x^3\]
(y + xy) dx + (x − xy2) dy = 0
(x + y) (dx − dy) = dx + dy
\[\frac{dy}{dx} = \frac{y}{x} + \sin\left( \frac{y}{x} \right)\]
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.
The decay rate of radium at any time t is proportional to its mass at that time. Find the time when the mass will be halved of its initial mass.
The normal to a given curve at each point (x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4), find its equation.
Radium decomposes at a rate proportional to the quantity of radium present. It is found that in 25 years, approximately 1.1% of a certain quantity of radium has decomposed. Determine approximately how long it will take for one-half of the original amount of radium to decompose?
The slope of the tangent at each point of a curve is equal to the sum of the coordinates of the point. Find the curve that passes through the origin.
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
Form the differential equation of the family of circles having centre on y-axis and radius 3 unit.
Determine the order and degree of the following differential equations.
| Solution | D.E |
| y = aex + be−x | `(d^2y)/dx^2= 1` |
Solve the following differential equation.
`y^3 - dy/dx = x dy/dx`
For the following differential equation find the particular solution.
`dy/ dx = (4x + y + 1),
when y = 1, x = 0
Solve the following differential equation.
`dy /dx +(x-2 y)/ (2x- y)= 0`
Solve the following differential equation.
`dy/dx + y` = 3
The solution of `dy/dx + x^2/y^2 = 0` is ______
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 the following differential equation `("d"y)/("d"x)` = x2y + y
Solve the following differential equation
`yx ("d"y)/("d"x)` = x2 + 2y2
Solve the following differential equation
`x^2 ("d"y)/("d"x)` = x2 + xy − y2
The function y = ex is solution ______ of differential equation
Solve the following differential equation
`y log y ("d"x)/("d"y) + x` = log y
Verify y = `a + b/x` is solution of `x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0
y = `a + b/x`
`(dy)/(dx) = square`
`(d^2y)/(dx^2) = square`
Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`
= `x square + 2 square`
= `square`
Hence y = `a + b/x` is solution of `square`
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
If `y = log_2 log_2(x)` then `(dy)/(dx)` =
