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we have (D^{3}  9D)y = cos x
implies D(D^{2}  9)y = cos x
So, complete solution is
Consider the differential equation y" + 6y' + 25y  0 with initial condition y(0) = 0. Then, the general solution of the IVP is
we have (D^{2} + 6D + 25)y = 0
So, y =e^{3x} [A cos 4x + B sin 4x]
but y(0) = 0 implies A = 0
Hence, y = Be^{3x} sin 4x
The solution of the differential equation y" + 4y = 0 subject to y(0) = 1, y' (0) = 2 is
we have (D^{2} + 4) = 0
So, solutions is y = c_{1} sin 2x + c_{2} cos 2x
but y(0)= 1 implies c_{2} = 1
and y'(0) = 2 implies 2c_{1}
= 2 implies c_{1} = 1
Hence y = sin 2x + cos 2x
General solution of the differential equation is given by
we have xdy = (y + xe^{y/x})dx
implies xdy  ydx = xe^{y/x} dx
Correct Answer : A
Explanation : y'' m^{2}y = 0
r^{2}  m^{2} = 0
r = +m
y = Ae^{m}x + Be^{m}x
Let A=C+D and B=DC
= Ce^{m}x + De^{m}x  Ce^{m}x + De^{m}x
= C(e^{m}xe^{m}x) + D(e^{m}x+e^{m}x)
= Csinh(mx) + Dcosh(mx)
The orthogonal trajectories of the curves y^{2} = 3x^{3}+x + c are
Differentiating w.r.t. x, we get
Replacing the equation of orthogonal trajectory is given by
implies
Integrating both sides, we get
implies 2 tan^{1} 3x + 3 In y = k
is satisfied, So, equation is exacct
...(i)
...(ii)
Adding (i) and (ii), writing the common term once.
The solution of the initial value problem xy'  y = 0 with y(1) = 1 is
If y'_{1} (x) = 3y_{1}(x) + 4y_{2}(x) and y'_{2} (x) = 4y_{1}(x) + 3y_{2}(x), then y_{1}(x) is
we have (D  3)y_{1}  4y_{2} = 0
4y_{2} + (3  D)y_{1} = 0
Solving equations, we get
So,
If e^{x} + xy + x sin y + e^{y} = c is the general solution of an exact differential equation, then the differential equation is
Taking total differential
we have
and
Since,
So, differential equation is exact
Now,
and
So, solution of differential equation
The differential equation (2x^{2} + by^{2})dx + cxydy = 0 is made exact by multiplying the integrating factor 1/x^{2}. Then the relation between and c is
Multiplying the differential equation by 1/x^{2}, we get
It is exact
So,
implies
2b + c = 0
If general solution of the differential equation ay'" + by" + cy’ + dy = 0 is linearly spanned by e^{x}, sinx and cosx, then which one of the following holds?
If y = In (sin (x + a)) + b, where a and b are constants, is the primitive, then the corresponding lowest order differential equation is
we have y = In sin (x + a ) + b
implies y'= cot (x + a)
implies y" =  cosec^{2} (x + a)
What is the degree of the nonhomogeneous partial differential equation,
Degree of an equation is defined as the power of the highest derivative present in the equation. Hence from the equation, the degree is 5.
Two linearly independent solutions of the differential equation y"  2y' + y = 0 are y_{1} = e^{x} and y_{2} = xe^{x}. Then a particular solution of y"  2y' + y = e^{x} sin x is
Consider the differential equation ( x + y + 1) dx + (2x + 2y + 1) dy = 0. Which of the following statements is true?
Which one of the following differential equations represents all circles with radius a?
(x  h)^{2} + (y  k)^{2} = a^{2} ...(i)
where h, k are parameters
Differentiating w.r.t. x, we get
( x  h ) + ( y  k ) y' = 0 ...(ii)
Differentiating w.r.t. x again 1 + ( y  k)
y^{11} + y^{12 } = 0
implies (iii)
From (ii), we get ( x  h ) =  ( y  k)y'
...(iv)
Putting (x  h) and (y  k) in (i), we get
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