Let k be real constant. The solution of the differential equations satisfies the relation
we have dy/dx = 2y + z and ...(i)
dz/dx = 3y ...(ii)
from 3 (i) l (ii), we get
implies
on integrating, we get
implies ln(3y + z) = 3x + c
So, 3y + z = ke^{3x}
Find the General Solution of the given differential equation.
This special case where D has a coefficient is solved using Legendre’s method.
Let log(8x + 7) = z
Then, e^{z} = 8x + 7
Substituting this in the equation,
8Dy + 2y = x
y(8D + 2) = x
Thus the auxiliary equation is 8D + 2 = 0
Thus, D = 1 / 4
(Solving by substituting powers of e^{z})
The General solution is C.F + P.I =
The differential equation 2ydx  (3y  2x)dy = 0
Here
The solution of the differential equation
we have
Consider the differential equation 2 cos (y^{2})dx  xy sin (y^{2})dy = 0
Correct Answer : D
Explanation : 2cos (y^{2})dx  xy sin (y^{2}) dy = 0
Multiplying by x^{3}
2x^{3} cos(y^{2})dx  x^{4}y sin(y^{2})dy = 0
dM/dy =  4x^{3}y sin(y^{2})
dN/dx =  4x^{3}y sin(y^{2})
Since dM/dy = dN/dx
So, x^{3} is an integrating factor.
Let f, g : [1, 1] → R, f (x) = x^{3}, g(x) = x^{2} IxI Then
y" + py' + qy = 0 on [1, 1]
Here, lxI = x for x < 0 = x for x ≥ 0
So, g(x) = x^{3} for x ≥ 0
= x^{3} for x < 0
and f(x) = x^{3}
If the general solutions of a differential equation are (y + c)^{2} = cx, where c is an arbitrary constant, then the order and degree of differential equation is
There will only one constant in the firstorder differential equation. Differentiating the given equation.
Putting the value of c in Eq. (1) and simplifying we will get a firstorder and seconddegree equation. Hence, (A) is the correct answer.
An integration factor of is
we have
The general solution of
(Here c_{1} and c_{2} are arbitrary constants.)
Let z = log x, Then
The differential equation changes to
implies
So,
Solution of the differential equation
=> x^{2} lnx dy  xy dy=xy dx – y^{2 }lny dx …….dividing by x^{2} y^{2} then
on integrating we get
where c is a constant of integration.
Let y_{1}(x) and y_{2}(x) be linearly independent solution of the differential equation y" + P(x) y' + Q(x)y = 0, where P(x) and Q(x) are continuous functions on an interval I. Then y_{3}(x) = ay_{1}(x) + by_{2}(x) and y_{4}(x) = cy_{1}(x) dy_{2(}x) are linearly independent solutions of the given differential equations if
y_{3}(x) and y_{4}(x) are linearly dependent if y_{3}(x) = ky_{4}(x) where k is any constant
Since, y_{1}(x) and y_{2}(x) are independent
So, a  kc = 0
and b  kd = 0
or
So, y_{3}(x) and y_{4}(x) are independent if
implies ad ≠ bc
Consider the differential equations, satisfying y(0) = 0, where This initial value problem
we have
or,
Thus,
implies y = x^{2} + c for x ≤ 0,
but y(0) = 0 implies c = 0
So,
The particular integral of the differential equation
y" + y' + 3y = 5 cos(2x + 3) is
we have (D^{2} + D + 3)y = 5 cos(2x + 3)
So,
The general solution of differential equation
4x^{2}y"  8xy' + 9y = 0 is
Let z = log x, Then
So, The differential equation becomes
So,
What is the order of the nonhomogeneous partial differential equation,
The order of an equation is defined as the highest order derivative present in the equation and hence from the equation,it is clear that it of 2^{nd} order.
If g(x, y)dx + (x + y )dy = 0 is an exact differential equation and if g(x, 0) = x^{2}, then the general solution of the differential equation is
g(x, y)dx + (x + y) dy = 0 is exact if
implies
g (x , 0) = x^{2}
So, g(x, y) = x^{2} + y
So, general solution is
implies 2x^{3} + 6xy + 3y^{2} = c
A general solution of the differential equation
Consider the differential equation Then is equal to
we have
implies
Consider the differential equation dy/dx = ay  by^{2}, where a, b > 0 and y(0) = y_{0}. As x → + ∝, the solution y (x) tends to
So, Equation becomes,
Orthogonal trajectories of the family of curves (x  1)^{2} + y^{2} + 2ax = 0 are the solution of the differential equation
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