Calculus 3.6: solutions of first order differential equations

3.3 Solution: To solve Equation 1, first solve the equations... Linear Equations 4.1 Definition A first-order linear differential equation has the form y0+ px y = qx.. Linear Equations 4

Infinite Series and Differential Equations

Nguyen Thieu Huy

Hanoi University of Science and Technology

3.3 Solution: To solve Equation (1), first solve the equations

3.4 Integrating Factors: In general, Equation (1) is not exact.

Definition

A function I (x , y ) 6= 0 is an integrating factor for (1) if the equation

I (x , y )[M(x , y )dx + N(x , y )dy ] = 0 (2)

is exact

A solution to (1) is obtained by solving the exact differential equation (2)

To find I (x , y ) we solve ∂(I (x ,y )M(x ,y ))∂y = ∂(I (x ,y )N(x ,y ))∂x

This can be done in some special situations, results are listed in the following:

3.4 Integrating Factors: In general, Equation (1) is not exact.

Definition

A function I (x , y ) 6= 0 is an integrating factor for (1) if the equation

I (x , y )[M(x , y )dx + N(x , y )dy ] = 0 (2)

is exact

A solution to (1) is obtained by solving the exact differential equation (2)

To find I (x , y ) we solve ∂(I (x ,y )M(x ,y ))∂y = ∂(I (x ,y )N(x ,y ))∂x

This can be done in some special situations, results are listed in the following:

3.4 Integrating Factors: In general, Equation (1) is not exact.

Definition

A function I (x , y ) 6= 0 is an integrating factor for (1) if the equation

I (x , y )[M(x , y )dx + N(x , y )dy ] = 0 (2)

is exact

A solution to (1) is obtained by solving the exact differential equation (2)

To find I (x , y ) we solve ∂(I (x ,y )M(x ,y ))∂y = ∂(I (x ,y )N(x ,y ))∂x

This can be done in some special situations, results are listed in the following:

Example: Solve ydx − xdy = 0 (not exact) Check

1

N



∂M(x ,y )

∂y −∂N(x ,y )∂x = −x2; Integrating factor I (x ) = e−R 2x dx = x12

The Eq ⇔ xy2dx −1xdy = 0 which is exact The solution is yx = C

This equation can be solved by the method described previously However,

we can proceed in a simpler way by rewriting it as

d (eR p(x)dxy )

R p(x)dxq(x )

⇔ eR p(x)dxy =R eR p(x)dxq(x )dx + C Thegeneral solution isy = e−R p(x)dxR eR p(x)dxq(x )dx + C∀ constant C

Example: Solve ydx − xdy = 0 (not exact) Check

1

N



∂M(x ,y )

∂y −∂N(x ,y )∂x = −x2; Integrating factor I (x ) = e−R 2x dx = x12

The Eq ⇔ xy2dx −1xdy = 0 which is exact The solution is yx = C

4 Linear Equations

4.1 Definition

A first-order linear differential equation has the form y0+ p(x )y = q(x )

Example Looking back: dvdt + kmv = g ; dtdI +RLI = EL

This equation can be solved by the method described previously However,

we can proceed in a simpler way by rewriting it as

d (eR p(x)dxy )

R p(x)dxq(x )

⇔ eR p(x)dxy =R eR p(x)dxq(x )dx + C Thegeneral solution is y = e−R p(x)dxR eR p(x)dxq(x )dx + C∀ constant C

Example: Solve ydx − xdy = 0 (not exact) Check

1

N



∂M(x ,y )

∂y −∂N(x ,y )∂x = −x2; Integrating factor I (x ) = e−R 2x dx = x12

The Eq ⇔ xy2dx −1xdy = 0 which is exact The solution is yx = C

4 Linear Equations

4.1 Definition

A first-order linear differential equation has the form y0+ p(x )y = q(x )

Example Looking back: dvdt + kmv = g ; dtdI +RLI = EL

(p(x )y − q(x ))dx + dy = 0 Check N1



∂M(x ,y )

∂y −∂N(x ,y )∂x = p(x )

Integrating factor: I (x ) = eR p(x)dx 6= 0 When multiplied by I (x), the

resulting equation eR p(x)dxy0+ eR p(x)dxp(x )y = eR p(x)dxq(x ) is exact

This equation can be solved by the method described previously However,

we can proceed in a simpler way by rewriting it as

d (eR p(x)dxy )

R p(x)dxq(x )

⇔ eR p(x)dxy =R eR p(x)dxq(x )dx + C Thegeneral solution is y = e−R p(x)dxR eR p(x)dxq(x )dx + C∀ constant C

Example: Solve ydx − xdy = 0 (not exact) Check

1

N



∂M(x ,y )

∂y −∂N(x ,y )∂x = −x2; Integrating factor I (x ) = e−R 2x dx = x12

The Eq ⇔ xy2dx −1xdy = 0 which is exact The solution is yx = C

4 Linear Equations

4.1 Definition

A first-order linear differential equation has the form y0+ p(x )y = q(x )

Example Looking back: dvdt + kmv = g ; dtdI +RLI = EL

(p(x )y − q(x ))dx + dy = 0 Check N1



∂M(x ,y )

∂y −∂N(x ,y )∂x = p(x )

Integrating factor: I (x ) = eR p(x)dx 6= 0 When multiplied by I (x), the

resulting equation eR p(x)dxy0+ eR p(x)dxp(x )y = eR p(x)dxq(x ) is exact

This equation can be solved by the method described previously However,

we can proceed in a simpler way by rewriting it as

d (eR p(x)dxy )

R p(x)dxq(x )

⇔ eR p(x)dxy =R eR p(x)dxq(x )dx + C Thegeneral solution is y = e−R p(x)dxR eR p(x)dxq(x )dx + C∀ constant C

Example: Solve ydx − xdy = 0 (not exact) Check

A first-order linear differential equation has the form y0+ p(x )y = q(x )

Example Looking back: dvdt + kmv = g ; dtdI +RLI = EL

we can proceed in a simpler way by rewriting it as

d (eR p(x)dxy )

R p(x)dxq(x ) ⇔ eR p(x)dxy =R eR p(x)dxq(x )dx + C Thegeneral solution is y = e−R p(x)dxR eR p(x)dxq(x )dx + C∀ constant C

Example Solve y0+4xy = x4 For p(x ) = 4x and q(x ) = x4, we obtain

If α > 0, then y = 0 is a solution Otherwise, if α < 0, then the condition

is y 6= 0 We now find the solutions y 6= 0

To do this, we divide both sides by yα to obtain y−αy0+ p(x )y1−α= q(x )

The substitution z = y1−α (so, z0 = (1 − α)y−αy0) now transforms Eq

into a linear differential equation z0+ (1 − α)p(x )z = (1 − α)q(x )

Example: Solve y0+ xy = xy2.This is a Bernoulli differential equation with p(x ) = q(x ) = x , and α = 2

First, y = 0 is a solution To find the solution y 6= 0 Dividing by y2 wehave y−2y0+ xy−1= x Subs z = y−1, ⇔ y = 1/z, then z0= −y−2y0.Subs into above equation: z0−xz =−x (linear Eq)

⇔ z = e− R (−x)dx R eR (−x)dx(−x)dx + C = 1 + C e 2 So, y = 1

1+C ex 22

Example Solve y0+4xy = x4 For p(x ) = 4x and q(x ) = x4, we obtain

If α > 0, then y = 0 is a solution Otherwise, if α < 0, then the condition

is y 6= 0 We now find the solutions y 6= 0

To do this, we divide both sides by yα to obtain y−αy0+ p(x )y1−α= q(x )

The substitution z = y1−α (so, z0 = (1 − α)y−αy0) now transforms Eq

into a linear differential equation z0+ (1 − α)p(x )z = (1 − α)q(x )

Example: Solve y0+ xy = xy2.This is a Bernoulli differential equation with p(x ) = q(x ) = x , and α = 2

First, y = 0 is a solution To find the solution y 6= 0 Dividing by y2 wehave y−2y0+ xy−1= x Subs z = y−1, ⇔ y = 1/z, then z0 = −y−2y0.Subs into above equation: z0−xz =−x (linear Eq)

⇔ z = e− R (−x)dx R eR (−x)dx(−x)dx + C = 1 + C e 2 So, y = 1

1+C ex 22

Example Solve y0+4xy = x4 For p(x ) = 4x and q(x ) = x4, we obtain

If α > 0, then y = 0 is a solution Otherwise, if α < 0, then the condition

is y 6= 0 We now find the solutions y 6= 0

To do this, we divide both sides by yα to obtain y−αy0+ p(x )y1−α = q(x )

The substitution z = y1−α (so, z0 = (1 − α)y−αy0) now transforms Eq

into a linear differential equation z0+ (1 − α)p(x )z = (1 − α)q(x )

Example: Solve y0+ xy = xy2.This is a Bernoulli differential equation with p(x ) = q(x ) = x , and α = 2

First, y = 0 is a solution To find the solution y 6= 0 Dividing by y2 wehave y−2y0+ xy−1= x Subs z = y−1, ⇔ y = 1/z, then z0 = −y−2y0.Subs into above equation: z0−xz =−x (linear Eq)

⇔ z = e− R (−x)dx R eR (−x)dx(−x)dx + C = 1 + C e 2 So, y = 1

1+C ex 22

Example Solve y0+4xy = x4 For p(x ) = 4x and q(x ) = x4, we obtain

If α > 0, then y = 0 is a solution Otherwise, if α < 0, then the condition

is y 6= 0 We now find the solutions y 6= 0

To do this, we divide both sides by yα to obtain y−αy0+ p(x )y1−α = q(x )

The substitution z = y1−α (so, z0 = (1 − α)y−αy0) now transforms Eq

into a linear differential equation z0+ (1 − α)p(x )z = (1 − α)q(x )

Example: Solve y0+ xy = xy2

This is a Bernoulli differential equation with p(x ) = q(x ) = x , and α = 2

First, y = 0 is a solution To find the solution y 6= 0 Dividing by y2 we

have y−2y0+ xy−1= x Subs z = y−1, ⇔ y = 1/z, then z0 = −y−2y0

Subs into above equation: z0−xz =−x (linear Eq)

⇔ z = e− R (−x)dx R eR (−x)dx(−x)dx + C = 1 + C e 2 So, y = 1

1+C ex 22

Example Solve y0+4xy = x4 For p(x ) = 4x and q(x ) = x4, we obtain

If α > 0, then y = 0 is a solution Otherwise, if α < 0, then the condition

is y 6= 0 We now find the solutions y 6= 0

To do this, we divide both sides by yα to obtain y−αy0+ p(x )y1−α = q(x ).The substitution z = y1−α (so, z0 = (1 − α)y−αy0) now transforms Eq.into a linear differential equation z0+ (1 − α)p(x )z = (1 − α)q(x )

Example: Solve y0+ xy = xy2

This is a Bernoulli differential equation with p(x ) = q(x ) = x , and α = 2.First, y = 0 is a solution To find the solution y 6= 0 Dividing by y2 wehave y−2y0+ xy−1= x Subs z = y−1, ⇔ y = 1/z, then z0 = −y−2y0.Subs into above equation: z0−xz =−x (linear Eq)

⇔ z = e− R (−x)dx

R eR (−x)dx(−x)dx + C= 1 + C ex 22 So, y = 1 x 2

VI Modeling:

Three main steps for modeling:

situation into mathematical terms:

(A) To choose the (physical, natural, ) laws governing the problem.

(B) To choose mathematical variables.

(C) To establish the (differential) equations.

Analysis of the Model

(A) To solve equations

(B) To find the properties of solutions

(C) If it is too difficult to solve the equations, try to approximate the solutions (numerical methods).

(A) To interpret the information from solutions (obtained above) in the context in which the problem arose.

(B) To check that the mathematical solution appears physically reasonable.

(C) To calculate the values of the solution at selected points and compare them with experimentally observed values Or, ask whether the behavior

of the solution after a long time is consistent with observations.

VI Modeling:

Three main steps for modeling:

situation into mathematical terms:

(A) To choose the (physical, natural, ) laws governing the problem.

(B) To choose mathematical variables.

(C) To establish the (differential) equations.

(A) To solve equations

(B) To find the properties of solutions

(C) If it is too difficult to solve the equations, try to approximate the

solutions (numerical methods).

(A) To interpret the information from solutions (obtained above) in the context in which the problem arose.

(B) To check that the mathematical solution appears physically reasonable.

(C) To calculate the values of the solution at selected points and compare them with experimentally observed values Or, ask whether the behavior

of the solution after a long time is consistent with observations.

VI Modeling:

Three main steps for modeling:

situation into mathematical terms:

(A) To choose the (physical, natural, ) laws governing the problem.

(B) To choose mathematical variables.

(C) To establish the (differential) equations.

(A) To solve equations

(B) To find the properties of solutions

(C) If it is too difficult to solve the equations, try to approximate the solutions (numerical methods).

(A) To interpret the information from solutions (obtained above) in the context in which the problem arose.

(B) To check that the mathematical solution appears physically reasonable.

(C) To calculate the values of the solution at selected points and compare them with experimentally observed values Or, ask whether the behavior

of the solution after a long time is consistent with observations.

Example RL-circuit:

1st Step: Modeling

Physical laws: 1) Kirchhoff’s voltage law (KVL): The algebraic sum of all

the instantaneous voltage drops around any closed loop is zero, or the

voltage impressed on a closed loop is equal to the sum of the voltage

drops in the rest of the loop.: VR + VL = E (t)

2) Ohm’s law: The voltage drop VR across a resistor is proportional to the

current I : VR = RI ;

inductor is proportional to the instantaneous time rate of change of the

Example RL-circuit:

1st Step: Modeling

Physical laws: 1) Kirchhoff’s voltage law (KVL): The algebraic sum of allthe instantaneous voltage drops around any closed loop is zero, or thevoltage impressed on a closed loop is equal to the sum of the voltagedrops in the rest of the loop.: VR + VL = E (t)

2) Ohm’s law: The voltage drop VR across a resistor is proportional to thecurrent I : VR = RI ;

inductor is proportional to the instantaneous time rate of change of thecurrent I: VL= LdIdt;

2nd Step: Solution to the equation dI

Case B: Periodic electromotive force E = E0sin ωt For this E (t) wehave that

2nd Step: Solution to the equation dI

Case B: Periodic electromotive force E = E0sin ωt For this E (t) wehave that

2nd Step: Solution to the equation dI

current I (t) will be nearly constant, equal to E0

R, the value it would haveimmediately (by Ohm’s law) had we no inductor in the circuit, and we see

that this limit is independent of the initial value I (0)

Case B: Periodic electromotive force E = E0sin ωt For this E (t) wehave that

2nd Step: Solution to the equation dI

R

LI = E (t)L Using formula

y = e−R p(x)dxR eR p(x)dxq(x )dx + Cwith x = t, y = I , p = R/L := α,and q = E /L we get

R, the value it would haveimmediately (by Ohm’s law) had we no inductor in the circuit, and we seethat this limit is independent of the initial value I (0)

Case B: Periodic electromotive force E = E0sin ωt For this E (t) wehave that

R2+ ω2L2sin(ωt − δ) where δ = arctan(ωL/R).

The exponential term will approach zero as t → ∞ This mean that aftersome time the current I (t) will execute practically harmonic oscillations

7 Existence and Uniqueness TheoremLet the functions f (x , y ) and ∂f∂y be continuous in some rectangle

α < x < β, γ < y < δ containing the point (x0, y0) Then, in someinterval (x0− h, x0+ h) contained in (α, β), there is a unique solution

y = ϕ(x ), x ∈ (x0− h, x0+ h), of the initial value problem

y0= f (x , y ), y (x0) = y0

R2+ ω2L2sin(ωt − δ) where δ = arctan(ωL/R).

The exponential term will approach zero as t → ∞ This mean that after

some time the current I (t) will execute practically harmonic oscillations

7 Existence and Uniqueness TheoremLet the functions f (x , y ) and ∂f∂y be continuous in some rectangle

α < x < β, γ < y < δ containing the point (x0, y0) Then, in someinterval (x0− h, x0+ h) contained in (α, β), there is a unique solution

y = ϕ(x ), x ∈ (x0− h, x0+ h), of the initial value problem

y0= f (x , y ), y (x0) = y0

R2+ ω2L2sin(ωt − δ) where δ = arctan(ωL/R).

The exponential term will approach zero as t → ∞ This mean that after

some time the current I (t) will execute practically harmonic oscillations

7 Existence and Uniqueness TheoremLet the functions f (x , y ) and ∂f∂y be continuous in some rectangle

α < x < β, γ < y < δ containing the point (x0, y0) Then, in someinterval (x0− h, x0+ h) contained in (α, β), there is a unique solution

y = ϕ(x ), x ∈ (x0− h, x0+ h), of the initial value problem

y0= f (x , y ), y (x0) = y0

R2+ ω2L2sin(ωt − δ) where δ = arctan(ωL/R).

The exponential term will approach zero as t → ∞ This mean that aftersome time the current I (t) will execute practically harmonic oscillations

7 Existence and Uniqueness Theorem

Let the functions f (x , y ) and ∂f∂y be continuous in some rectangle

α < x < β, γ < y < δ containing the point (x0, y0) Then, in some

interval (x0− h, x0+ h) contained in (α, β), there is a unique solution

y = ϕ(x ), x ∈ (x0− h, x0+ h), of the initial value problem

y0= f (x , y ), y (x0) = y0