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This mock test of Test: Linear Differential Equations for JEE helps you for every JEE entrance exam.
This contains 10 Multiple Choice Questions for JEE Test: Linear Differential Equations (mcq) to study with solutions a complete question bank.
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QUESTION: 1

The solution of the differential equation is :

Solution:

dy/dx + 3y = 2e^{3}x

p = 3, q = 2e^{3}x

∫p.dx = 3x

Integrating factor (I.F) = e^{3}x

y(I.F) = ∫Q(I.F) dx

ye^{3}x = ∫2e^{3}x e^{3}x dx

ye^{3}x = ∫2e^{6}x dx

ye^{3}x = 2∫e^{6}x dx

ye^{3}x = 2/6[e^{6}x] + c

ye^{3}x = ⅓[e^{6}x] + c

Dividing by e^{3}x, we get

y = ⅓[e^{3}x] + ce^{-(3x)}

QUESTION: 2

The integrating factor of differential equation is :

Solution:

xlog x dy/dx + y = 2logx

⇒ dy/dx + y/(xlogx) = 2/x...(1)

Put P = 1/x logx

⇒∫PdP = ∫1/x logx dx

= log(logx)

∴ I.F.= e^{∫PdP} = e^{log}(logx)

= logx

QUESTION: 3

The solution of the differential equation x dy = (2y + 2x^{4} + x^{2}) dx is:

Solution:

xdy = (2y + 2x^{4 }+ x^{2})dx

→ dy/dx − (2x)y = 2x^{3} + x

This differential is of the form y′+P(x)y=Q(x) which is the general first order linear differential equation, where P(x) and Q(x) are continuous function defined on an interval.

The general solution for this is y∙I.F = ∫I.F × Q(x)dx

Where I.F = e∫P(x)dx is the integrating factor of the differential equation.

I.F = e∫P(x)dx

= e^{∫−2}/xdx

= e(−2∙lnx)

= e^{ln}(x^{−2})

= x^{−2}

Thus y(1/x^{2}) = ∫1/x^{2}(2x^{3} + x)dx

=∫(2x + 1/x)dx

= x^{2} + lnx + C

⟹ y = x^{4} + x^{2}lnx + c

QUESTION: 4

The solution of the differential equation

Solution:

QUESTION: 5

The integrating factor of differential equation is :

Solution:

dy/dx+2ytanx=sinx

This is in the form of dy/dx + py = θ

where p=2tanx,θ=sinx

∴ finding If e^{∫pdx} = e^{∫2tanxdx }

=e^{(2log secx)}

=e^{(log sec2x)}

=sec^{2}x

QUESTION: 6

The solution of the differential equation is :

Solution:

dy/dx + y/x = x^{2}

differential equation is in the form : dy/dx + Py = Q

P = 1/x Q = x^{2}

I.F = e∫P(x)dx

I.F = e∫1/x dx

I.F = e[log x]

I.F = x

y I.F = ∫(Q * I.F) dx + c

yx = ∫x^{2} * x * dx + c

yx = ∫x^{3} dx + c

xy = [x^{4}]/4 + c

QUESTION: 7

The solution of the differential equation is :

Solution:

QUESTION: 8

The solution of the differential equation is :

Solution:

The given differential equation may be written as

dy/dx + (2x/(x^{2}+1)y = √x2+4/(x^{2}+1) ... (i)

This is of the form dy/dx + Py = Q,

where P=2x/(x^{2}+1) and Q=(√x^{2}+4)/(x^{2}+1)

Thus, the given differential equation is linear.

IF=e^{(∫Pdx)}

= e(∫2x(x^{2}+1)dx)

= e(log(x^{2}+1) = (x^{2}+1)

So, the required solution is given by

y × IF = ∫{Q×IF}dx + C,

i.e., y(x^{2}+1)=∫(√x^{2}+4)/(x^{2}+1)×(x^{2}+1)dx

⇒y(x^{2}+1)=∫(√x^{2}+4)dx

=1/2x (√x^{2}+4) +1/2 × (2)^{2} × log|x+(√x^{2}+4)| + C

=1/2x (√x^{2}+4) + 2log|x+(√x^{2}+4) + C.

Hence, y(x^{2}+1) = 1/2x(√x^{2}+4) + 2log|x+√x^{2}+4| + C is the required solution.

QUESTION: 9

The integrating factor of differential equation is :

Solution:

QUESTION: 10

The integrating factor of differential equation is :

Solution:

dy/dx + y/x = x

Differential eq is in the form of : dy/dx + Py = Q

P = 1/x Q = x

I.F. = e^{∫Pdx}

= e^{∫(1/x)}dx

= e^{(log x)}

⇒ x

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