Differential Equations - 12 - Mathematics MCQ

Differentiating equation (i) w.r.t. we get

Linearly independent solutions of ay" + by”+cy' + dy = 0 are e^x cos x and sin x, so roots of aD³ + bD² + cD + d = 0 will be 1 and ±f, so we have 1 + 

Particular integral of y" - 2y' + y - e^x sin x is

By putting the value of a from equation (ii) in equation (i), we get.


It ’s orthogonal trajectory will be D.E.

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

Integrating factor of (2x² + by²) dx + cxy dy = 0 is 

 ....(i)
Differentiating both sides w.r.t. x, we get.

Equation of tangent is (Y - y) = y'(X - x)

Intercept on Y-axis.

Intercept on X-axis:

0
 

Correct Answer :- C

Explanation : dy/dx = (x+y-z)/(x+y)

Let t = x + y

1 + dy/dx = dt/dx

Equation becomes,

dt/dx - 1 = (t-z)/t

dt/dx = (2t-z)/t

tdt/(2t-z) = dx

= 1/2[1+z/(2t-z)]dt = dx

t + 1/2log(2t-z) = 2x+c

2t-z = e2^(x-y+c)

x+y-1 = ce^x-y

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

Correct Answer :- D

Explanation : y(x²y + e^x)dx - e^xdy = 0

Dividing it by y²

(x² + e^x/y)dx - e^x/y² dy = 0

x² dx + (ye^x - e^x dy)/y² = 0

d(x³/3) + d(e^x/y) = 0

∫d(x³/3) + ∫d(e^x/y) = 0

=> (x³/3) + (e^x/y) = c

=> x³y + 3e^x = 3yc