Advertisements
Advertisements
Question
Solve
`dy/dx + 2/ x y = x^2`
Solution
`dy/dx + 2/ x y = x^2`
The given equation is of the form
`dy/dx + py = Q`
`where, P = 2/x and Q = x^2`
∴ I.F. =`e^(int^(pdx) = e^(2int^(1/xdx) e = ^(2logx) = e^(logx^2) = x^2`
∴ Solution of the given equation is
`y(I.F.) = int Q(I.F.) dx + c_1`
`y(x^2) = int x^2 xx x^2 dx + c_1`
∴ `x ^2 y = x^4 intdx + c_1`
∴ `x^2 y = x^5/5 + c_1`
∴ 5x2 y = x5 + c …[c = 5c1]
APPEARS IN
RELATED QUESTIONS
Verify that y = − x − 1 is a solution of the differential equation (y − x) dy − (y2 − x2) dx = 0.
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x\frac{dy}{dx} + y = y^2\]
|
\[y = \frac{a}{x + a}\]
|
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
(y + xy) dx + (x − xy2) dy = 0
Solve the following differential equation:
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]
x2 dy + y (x + y) dx = 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.
Write the differential equation representing the family of straight lines y = Cx + 5, where C is an arbitrary constant.
Which of the following transformations reduce the differential equation \[\frac{dz}{dx} + \frac{z}{x}\log z = \frac{z}{x^2} \left( \log z \right)^2\] into the form \[\frac{du}{dx} + P\left( x \right) u = Q\left( x \right)\]
Find the differential equation whose general solution is
x3 + y3 = 35ax.
For each of the following differential equations find the particular solution.
`y (1 + logx)dx/dy - x log x = 0`,
when x=e, y = e2.
Solve the following differential equation.
`xy dy/dx = x^2 + 2y^2`
Solve: `("d"y)/("d"x) = cos(x + y) + sin(x + y)`. [Hint: Substitute x + y = z]
A man is moving away from a tower 41.6 m high at a rate of 2 m/s. If the eye level of the man is 1.6 m above the ground, then the rate at which the angle of elevation of the top of the tower changes, when he is at a distance of 30 m from the foot of the tower, is
`d/(dx)(tan^-1 (sqrt(1 + x^2) - 1)/x)` is equal to:
