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The solution of the differential equation yy' + y^{2}  x = 0, where c is a constant, is
Let k be a real constant. The solution of the differential equations satisfies the relation
Differentiating equation (i) w.r.t. we get
If general solution of the differential equation ay",+ by" cy’+ dy = 0 is linearly spanned by e^{x} sin x and cos x, then which one of the following holds?
Linearly independent solutions of ay" + by”+cy' + dy = 0 are e^{x} cos x and sin x, so roots of aD^{3} + bD^{2} + cD + d = 0 will be 1 and ±f, so we have 1 +
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^{2}  2y' + y + e^{x} sin x is
Particular integral of y"  2y' + y  e^{x} sin x is
Orthogonal trajectories of the family of curves (x  1)^{2} + y^{2} + 2ax = 0 are the solutions of the differential equation
By putting the value of a from equation (ii) in equation (i), we get.
It ’s orthogonal trajectory will be D.E.
Which one of the following differential equations represents all circles with radius a?
Equation of the circles with radius ‘a' is
Putting values of (y  k) and (x  h) from equation (iv) and equation (v) in equation (i), we get
The solution of the differential equation with initial condition y(0) = 0 is
The differential equation (2a^{2} + by^{2}) dx + cxydy = 0 is made exact by multiplying the integrating factor Then the relation between b and c is.
Integrating factor of (2x^{2} + by^{2}) dx + cxy dy = 0 is
....(i)
Differentiating both sides w.r.t. x, we get.
The sum of the intercepts made on the axes of coordinates by any tangent to the curve √x + √y = 2 is equal to
Equation of tangent is (Y  y) = y'(X  x)
Intercept on Yaxis.
Intercept on Xaxis:
0
If and it is known that for a = 1, y = 1; if x = 1, then the value of y will be :
If the solution of the differential equation is x + y 1 = Ce^{u} then the value of u is :
Correct Answer : C
Explanation : dy/dx = (x+yz)/(x+y)
Let t = x + y
1 + dy/dx = dt/dx
Equation becomes,
dt/dx  1 = (tz)/t
dt/dx = (2tz)/t
tdt/(2tz) = dx
= 1/2[1+z/(2tz)]dt = dx
t + 1/2log(2tz) = 2x+c
2tz = e2^{(xy+c)}
x+y1 = ce^{xy}
which is linear differential equation so, it’s solution is
which is linear differential equation in dependent variable x and independent variable z.
whose solution is
.......(i)
which is linear differential equation, so solution is
The degree and order of differential equation are respectively
The solution of the equation (2x + y + 1)dx + (4x + 2y – 1)dy = 0 is:
The solution of the differential equation y(x^{2}y + e^{x})dx e^{x}dy = 0 is
Correct Answer : D
Explanation : y(x^{2}y + e^{x})dx  e^{x}dy = 0
Dividing it by y^{2}
(x^{2} + e^{x}/y)dx  e^{x}/y^{2} dy = 0
x^{2} dx + (ye^{x}  e^{x} dy)/y^{2} = 0
d(x^{3}/3) + d(e^{x}/y) = 0
∫d(x^{3}/3) + ∫d(e^{x}/y) = 0
=> (x^{3}/3) + (e^{x}/y) = c
=> x^{3}y + 3e^{x} = 3yc
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