Solve the differential equation d y d x + y x = x 2 .

Given differential equation is, d y d x + y x = x 2 Which is linear differential equation of first order and is of the form, d y d x + P y = Q Where P = 1 x and Q = x 2 ∴ I . F = e ∫ P d x = e ∫ 1 x d x = e log x = x Multiply given equation by x , we get, x d y d x + y = x 3 ⇒ x y = x 4 4 + c ∴ y = x 3 4 + c x

Find the order and the degree of the differential equation: x 2 d 2 y d x 2 = 1 + d y d x 2 4 .

Given differential equation is x 2 d 2 y d x 2 = 1 + d y d x 2 4 ⇒ x 2 d 2 y d x 2 1 = 1 + d y d x 2 4 We know that, The order of a differential equation is the order of the highest derivative occurring in the differential equation. The degree of a differential equation is the power of the highest order derivative when differential coefficients are made free from radicals and fractions. Hence, Order = 2 and Degree = 1 .

Solve the differential equation d y d x = x log x

Given d y d x = x log x The above equation can be written as d y = x log x d x Integrate both sides ⇒ ∫ d y = ∫ x log x d x ⇒ y = log x ∫ x d x - ∫ ∫ x d x d d x log x d x (Using integration by part) ⇒ y = x 2 2 log x - ∫ x 2 2 1 x d x ⇒ y = x 2 2 log x - x 2 4 + C This is the required solution for the given differential equation.

Solve the differential equation: x d y - y d x = x 2 + y 2 d x , given that y = 0 when x = 1 .

Given the differential equation is x d y - y d x = x 2 + y 2 d x ⇒ x d y = x 2 + y 2 + y d x ⇒ d y d x = x 2 + y 2 + y x → 1 Here the numerator degree and denominator degree are having the same. It is a homogeneous differential equation Put y = v x and d y d x = v + x d v d x (Apply product rule in differentiation) Now put these values in 1 , we get ⇒ v + x d v d x = x 2 + v 2 x 2 + v x x = 1 + v 2 + v ⇒ x d v d x = 1 + v 2 ⇒ d v 1 + v 2 = d x x ⇒ ∫ d v 1 + v 2 = ∫ d x x (Apply integration) ⇒ log v + 1 + v 2 = log x + log C = log C x ⇒ v + 1 + v 2 = C x ⇒ y x + 1 + y 2 x 2 = C x Put y = v x ⇒ y + x 2 + y 2 = C x 2 → 2 When x = 1 , y = 0 (given), substitute these values in 2 , we get ⇒ 0 + 1 + 0 = C × ( 1 ) 2 ⇒ C = 1 ∴ y + x 2 + y 2 = x 2 is the required solution.

Solve the differential equation 1 + x 2 d y d x + 2 x y - 4 x 2 = 0 , subject to the initial condition y ( 0 ) = 0 .

Given differential equation 1 + x 2 d y d x + 2 x y - 4 x 2 = 0 , y ( 0 ) = 0 ⇒ 1 + x 2 d y d x + 2 x y = 4 x 2 Divide by 1 + x 2 on both sides, we get ⇒ d y d x + 2 x 1 + x 2 · y = 4 x 2 1 + x 2 It is of the form d y d x + P y = Q Here, P = 2 x 1 + x 2 , Q = 4 x 2 1 + x 2 Now, integrating factor I . F = e ∫ P d x ⇒ I . F = e ∫ 2 x 1 + x 2 d x ⇒ I . F = e log 1 + x 2 ∵ ∫ f ' ( x ) f x d x = log f x + C ⇒ I . F = 1 + x 2 ∴ The general solution is y × I . F = ∫ Q × I . F d x + C ⇒ y · 1 + x 2 = ∫ 4 x 2 1 + x 2 × 1 + x 2 d x + C ⇒ y · 1 + x 2 = 4 ∫ x 2 d x + C ⇒ y · 1 + x 2 = 4 x 3 3 + C → 1 Now put x = 0 , y = 0 in 1 , we get ⇒ C = 0 Put the value of C in 1 , we get ⇒ y · 1 + x 2 = 4 x 3 3 .

Solve the differential equation d y d x + y = e - 2 x

Given differential equation is d y d x + y = e - 2 x . It is a linear differential equation in y . Comparing it with d y d x + P y = Q , We get P = 1 , Q = e - 2 x Integrating factor IF = e ∫ P d x = e ∫ d x = e x So, the integrating factor, IF = e x The general solution is y · IF = ∫ Q · IF d x ⇒ y · e x = ∫ e - 2 x · e x d x ⇒ y e x = ∫ e - x d x ⇒ y e x = - e - x + c ⇒ y = - e - 2 x + c e - x which is the required solution of given differential equation.

Solve the differential equation 4 d y d x + 8 y = 5 e - 3 x

The given equation is 4 d y d x + 8 y = 5 e - 3 x ⇒ d y d x + 2 y = 5 4 e - 3 x This is a linear differential equation and comparing it with d y d x + P y = Q we get, P = 2 , Q = 5 4 e - 3 x The integrating factor is I F = e ∫ P d x = e ∫ 2 d x = e 2 x The solution is given by, y = ∫ Q × I F d x + C I F where C is an arbitrary constant. Hence we get, y = ∫ 5 4 e - x d x + C e 2 x ⇒ y = - 5 e - x 4 + C e 2 x ⇒ y = - 5 e - 3 x 4 + C e - 2 x is the required solution.

Solve the differential equation d y d x + 2 y = e 3 x

The given equation is d y d x + 2 y = e 3 x This is a linear differential equation and comparing it with d y d x + P y = Q we get, P = 2 , Q = e 3 x The integrating factor is I F = e ∫ P d x = e ∫ 2 d x = e 2 x The solution is given by, y = ∫ Q × I F d x + C I F where C is an arbitrary constant. Hence we get, y = ∫ e 5 x d x + C e 2 x ⇒ y = e 5 x 5 + C e 2 x ⇒ y = e 3 x 5 + C e - 2 x is the required solution.

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Important Questions on Differential Equations

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Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

Solve

(1 + y^{2}) d x = (\tan^{- 1} y - x) d y

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

\frac{d y}{d x} + \frac{3}{\cos^{2} x} y = \frac{1}{\cos^{2} x}

x \in (- \frac{π}{3}, \frac{π}{3}),

and

y (\frac{π}{4}) = \frac{4}{3},

then

y (- \frac{π}{4})

equals

EASY

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

Solve the differential equation

\frac{d y}{d x} + \frac{y}{x} = x^{2} .

EASY

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

Find the order and the degree of the differential equation:

x^{2} \frac{d^{2} y}{d x^{2}} = {\{1 + {(\frac{d y}{d x})}^{2}\}}^{4}

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

The solution of

\frac{d y}{d x} + y = e^{- x}, y (0) = 0

is :

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

A $10 V$ cell of negligible internal resistance is connected in parallel across a battery of emf $200 V$ and internal resistance $38 Ω$ as shown in the figure. Find the value of current in the circuit.

Question Image

EASY

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

Solve the differential equation

\frac{d y}{d x} = x \log x

EASY

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

A parallel plate capacitor of capacitance

20 μ F

is being charged by a voltage source whose potential is changing at the rate of

3 V s^{- 1}

. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor would be, respectively:

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

Find the particular solution of the differential equation

e^{x} \tan y d x + (2 - e^{x}) \sec^{2} y d y = 0

, given that

y = \frac{π}{4}

when

x = 0

HARD

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

Solve the differential equation:

x d y - y d x = \sqrt{x^{2} + y^{2}} d x

, given that

y = 0

when

x = 1

EASY

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

The order and the degree of the differential equation

{(\frac{d^{3} y}{d x^{3}})}^{\frac{1}{6}} - {(\frac{d y}{d x})}^{\frac{1}{3}} = 0

are respectively

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

(2 + \sin x) \frac{d y}{d x} + (y + 1) \cos x = 0

and

y (0) = 1

, then

y (\frac{π}{2})

is equal to

HARD

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

Solve the differential equation

(1 + x^{2}) \frac{d y}{d x} + 2 x y - 4 x^{2} = 0

, subject to the initial condition

y (0) = 0

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

A tangent to the curve,

y = f (x)

P (x, y)

meets

x

-axis at

A

and

y

-axis at

B

. If

A P : B P = 1 : 3

and

f (1) = 1,

then the curve also passes through the point

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Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

y (x)

is the solution of the differential equation

(x + 2) \frac{d y}{d x} = x^{2} + 4 x - 9, x \neq - 2

and

y (0) = 0,

then

y (- 4)

is equal to

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

Solve the differential equation $\frac{d y}{d x} + y = e^{- 2 x}$

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

Solve the differential equation $4 \frac{d y}{d x} + 8 y = 5 e^{- 3 x}$

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

Solve the differential equation $\frac{d y}{d x} + 2 y = e^{3 x}$

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Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

Solve the following initial value problem

x (x^{2} + 3 y^{2}) d x + y (y^{2} + 3 x^{2}) d y = 0, y (1) = 1

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

The solution of the differential equation

\frac{d y}{d x} + x y = {x y}^{2}

is: (where

c

is an arbitrary constant)

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Important Questions on Differential Equations

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