Verify that y = e m cos - 1 x satisfies the differential equation 1 - x 2 d 2 y d x 2 - x d y d x - m 2 y = 0 .

Given y = e m cos - 1 x . . . . . . . i Differentiate y with respect to x , we get ⇒ d y d x = d d x e m cos - 1 x ⇒ d y d x = e m cos - 1 x d d x m cos - 1 x (Use chain rule) ⇒ d y d x = y - m 1 - x 2 . . . . . . . ii (From i ) Differentiate it again with respect to x , we get ⇒ d 2 y d x 2 = 1 - x 2 - m d y d x - ( - m y ) d d x 1 - x 2 1 - x 2 2 ⇒ d 2 y d x 2 = 1 - x 2 - m d y d x - ( - m y ) - 2 x 2 1 - x 2 1 - x 2 ⇒ d 2 y d x 2 = - 1 - x 2 m d y d x - ( m y ) x 1 - x 2 1 - x 2 ⇒ d 2 y d x 2 = - m 1 - x 2 d y d x - x m y 1 1 - x 2 1 - x 2 ⇒ d 2 y d x 2 = ( - m ) ( - m y ) + x d y d x 1 - x 2 ⇒ 1 - x 2 d 2 y d x 2 = m 2 y + x d y d x ⇒ 1 - x 2 d 2 y d x 2 - x d y d x - m 2 y = 0 Hence, y = e m cos - 1 x is a solution of 1 - x 2 d 2 y d x 2 - x d y d x - m 2 y = 0 .

Solve the differential equation x 3 + x 2 + x + 1 d y d x = 2 x 2 + x , y = 1 when x = 0 .

Given, x 3 + x 2 + x + 1 d y d x = 2 x 2 + x ⇒ d y d x = 2 x 2 + x x 3 + x 2 + x + 1 ⇒ d y = 2 x 2 + x x 2 + 1 x + 1 d x Integrating both sides, ⇒ ∫ d y = ∫ 2 x 2 + x x 2 + 1 x + 1 d x … i Using partial fraction, 2 x 2 + x x 2 + 1 x + 1 = A x + 1 + B x + C x 2 + 1 = A x 2 + 1 + B x + C x + 1 x 2 + 1 x + 1 ⇒ 2 x 2 + x = A x 2 + 1 + B x + C x + 1 Comparing both sides, A + B = 2 , B + C = 1 & A + C = 0 Solving them we get, A = 1 2 , B = 3 2 & C = - 1 2 Therefore, from equation i we get, y = ∫ 1 2 x + 1 + 3 2 x - 1 2 x 2 + 1 d x = 1 2 ∫ 1 x + 1 d x + 3 2 ∫ x x 2 + 1 d x - 1 2 ∫ 1 x 2 + 1 d x ⇒ y = 1 2 log x + 1 + 3 4 log x 2 + 1 - 1 2 tan - 1 x + c … i i Given, y = 1 when x = 0 ∴ i i becomes 1 = 1 2 log 0 + 1 + 3 4 log 0 + 1 - 1 2 tan - 1 0 + c ⇒ 1 = 0 + 0 - 0 + c ⇒ c = 1 Putting this value in equation i i y = 1 2 log x + 1 + 3 4 log x 2 + 1 - 1 2 tan - 1 x + 1 , which is the required equation.

Solve the following linear differential equation: x 2 + 1 d y d x + 2 x y = x 2 + 4 .

Given differential equation: x 2 + 1 d y d x + 2 x y = x 2 + 4 Here, to make coefficient of d y d x unity, so we divide both sides by x 2 + 1 , we get d y d x + 2 x y x 2 + 1 = x 2 + 4 x 2 + 1 → 1 Compare equation 1 with d y d x + P y = Q , then we have P = 2 x x 2 + 1 and Q = x 2 + 4 x 2 + 1 Now, the integrating factor I . F = e ∫ P d x i.e, I . F = e ∫ 2 x x 2 + 1 d x ⇒ I . F = e log x 2 + 1 ⇒ I . F = x 2 + 1 Therefore, the general solution of linear differential equation is y ( I . F ) = ∫ ( I . F ) Q d x + c ⇒ y x 2 + 1 = ∫ x 2 + 4 x 2 + 1 × x 2 + 1 d x + c ⇒ y x 2 + 1 = ∫ x 2 + 4 d x + c Using formula ∫ a 2 + x 2 d x = x 2 a 2 + x 2 + a 2 2 log x + a 2 + x 2 + C ∴ y x 2 + 1 = x 2 x 2 + 4 + 4 2 log x + x 2 + 4 + c ⇒ y x 2 + 1 = x 2 x 2 + 4 + 2 log x + x 2 + 4 + c , which is the required general solution.

Solve the differential equation ( x d y - y d x ) y sin y x = ( y d x + x d y ) x cos y x .

Given, ( x d y - y d x ) y sin y x = ( y d x + x d y ) x cos y x ⇒ x y sin y x d y - y 2 sin y x d x = x y cos y x d x + x 2 cos y x d y ⇒ x y sin y x - x 2 cos y x d y = x y cos y x + y 2 sin y x d x ⇒ d y d x = x y cos y x + y 2 sin y x x y sin y x - x 2 cos y x ⇒ d y d x = y x cos y x + y x 2 cos y x y x sin y x - cos y x ⇒ y = v x ⇒ d y d x = x d v d x + v ⇒ v + xdv dx = vx x cos vx x + v 2 x 2 x 2 sin vx x vx x sin vx x - cos vx x ⇒ x dv dx = vcosv + v 2 sinv vsinv - cosv - v ⇒ x dv dx = vcosv + v 2 sinv - v 2 sinv + vcosv vsinv - cosv ⇒ xdv dx = 2 vcosv vsinv - cosv ⇒ vsinv - cosv vcosv dv = 2 dx x ⇒ vsinv vcosv - cosv vcosv dv = 2 dx x ⇒ tanv - 1 v dv = 2 dx x ⇒ ∫ tanv - 1 v dv = 2 ∫ dx x ⇒ ∫ tanvdv - ∫ dv v = 2 ∫ dx x ⇒ log | secv | - log | v | = 2 log | x | + logC ⇒ log secv v = logx 2 + logC ⇒ log secv v = logx 2 c ⇒ secv v = x 2 c ⇒ sec y x y x = x 2 c ⇒ sec y x = y x x 2 c ⇒ sec y x = Cxy is the required differential equation.

Let T denote a curve y = y x which is in the first quadrant and let the point 1 , 0 lie on it. Let the tangent to T at a point P intersect the y -axis at Y P . If P Y P has length 1 for each point P on T , then which of the following options is/are correct?

y = - l o g e 1 + 1 - x 2 x + 1 - x 2

E M B I B E

Gamma Question Bank for Engineering Mathematics>Chapter 30 - Differential Equations>Assignment Set-1>Q 33

MEDIUM

JEE Main/Advance

IMPORTANT

Earn 100

It is known that, if the interest is compounded continuously, the principal changes at the rate equal to the product of the rate of bank interest per annum and the principal.

At what interest rate will $Rs . 100$ double itself in $10$ years?

Important Questions on Differential Equations

MEDIUM

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 30 - Differential Equations>Assignment Set-1>Q 33

It is known that, if the interest is compounded continuously, the principal changes at the rate equal to the product of the rate of bank interest per annum and the principal.

How much will $\Rightarrow 10 \times \log 2 = t$ be worth at $5 %$ interest after $10$ years? $\log 2 = 0.6931$ .

HARD

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 30 - Differential Equations>Assignment Set-1>Q 34

Verify that

y = e^{m \cos^{- 1} x}

satisfies the differential equation

(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - m^{2} y = 0

.

MEDIUM

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 30 - Differential Equations>Assignment Set-1>Q 34

Verify that

$(x^{2} - y^{2}) = c {(x^{2} + y^{2})}^{2}$ satisfies the differential equation $(x^{3} - 3 x y^{2}) d x = (y^{3} - 3 x^{2} y) d y$

HARD

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 30 - Differential Equations>Assignment Set-1>Q 35

A radioactive substance disintegrates at a rate proportional to the amount of substance present. If

50 %

of the given amount disintegrates in

1600

years. What percentage of the substance disintegrates in

10

years?

MEDIUM

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 30 - Differential Equations>Assignment Set-1>Q 36

Find the curve in

x - y

plane passing through the point

(1, 1)

so that the segment of any tangent drawn to the curve between its point of tangency and the

y

-axis is bisected at the

x

-axis.

HARD

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 30 - Differential Equations>Assignment Set-1>Q 37

A wet porous substance in the open air loses its moisture at a rate proportional to the moisture content. If a sheet hung in the wind loses half of its moisture during the first hour, when will it have lost