Linear Algebra and Differential Equations

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Linear Algebra and Differential Equations

Sujin Khomrutai, Ph.D.

Department of Math & Computer Science Chulalongkorn University

Lecture 2

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Wronskian & General Solutions

Theorem (General solutions) If y₁,y₂ are solutions of the ODE

y⁰⁰+p(t)y⁰+q(t)y = 0, and W(y1,y2)(t)6= 0 for all t, then the solution formula

y(t) =C₁y₁(t) +C₂y₂(t) is a general solution of the ODE.

On the other hand, if the Wronskian W(y₁,y₂)(t) = 0, then the formula y(t) =C1y1(t) +C2y2(t) is not a general solution.

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Wronskian & General Solutions

Proof. We have to show that for any k,l, one can chooseC₁,C₂ so that y(t) satisfies

y(t₀) =k, y⁰(t₀) =l.

This is the same as

y₁(t₀) y₂(t₀) y₁⁰(t0) y₂⁰(t0)

C₁ C2

= k

l

has solutionsC1,C2. But then we need thatW(y1,y2)(t0)6= 0 which is true by assumption.

Thus y(t) =C₁y₁(t) +C₂y₂(t) is a general solution of the ODE.

IfW(y1,y2)(t0) = 0, then [y1(t0),y₁⁰(t0)]^T and [y2(t0),y₂⁰(t0)]^T are linearly dependent, so there are no C₁,C₂ satisfying the equation when [k,l]^T is a non-zero vector perpendicular to [y₁(t₀),y₁⁰(t₀)]^T.

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Wronskian & General Solutions

EX.For the ODE y⁰⁰+ 5y⁰+ 6y = 0, show thaty₁=e^−2t andy₂ =e^−3t are solutions andy(t) =C1y1(t) +C2y2(t) is a general solution.

Sol.

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Wronskian & General Solutions

EX.For the ODE y⁰⁰+ 2y⁰+y = 0 show that y₁ =e^−t andy₂ =te^−t are solutions and y(t) =C1y1(t) +C2y2(t) is a general solution.

Sol.

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Wronskian & General Solutions

EX.Determine whether y(t) =C₁e^2t+C₂5e^2t is a general solution of the ODEy⁰⁰−y⁰−2y = 0?

Sol.

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Exercise

EX.For the ODE y⁰⁰−2y⁰−15y = 0, show thaty₁=e^−3t andy₂=e^5t are solutions and determine whether y(t) =C1y1(t) +C2y2(t) is a general solution.

Sol.

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Solutions to IVP

Definition (Fundamental set of solutions)

A pair of solutions y1,y2 toy⁰⁰+p(t)y⁰+q(t)y= 0 such that W(y1,y2)(t)6= 0 for all t

is called afundamental set of solutions to the ODE.

Theorem (Solutions to IVP) For the IVP

y⁰⁰+p(t)y⁰+q(t)y = 0, y(t₀) =k, y⁰(t₀) =l if y₁,y₂ is a fundamental set of solutions to the ODE, then

y(t) = det

k y₂(t₀) l y₂⁰(t0)

W(y₁,y₂)(t₀) y₁(t) + det

y₁(t₀) k y₁⁰(t0) l

W(y₁,y₂)(t₀) y₂(t).

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Solutions to IVP

EX.For the ODE

y⁰⁰+y⁰ = 0

show that y₁= 1 andy₂ =e^−t form a fundamental set of solutions and solve the IVP

y⁰⁰+y⁰ = 0, y(0) =−1, y⁰(0) = 2.

Sol.

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Constant Coefficients ODE

Definition ODE of the form

ay⁰⁰+by⁰+c =f(t)

is called asecond order constant coefficient equation.

Iff(t) = 0, the ODE is homogeneous.

Iff(t)6= 0, the ODE is nonhomogeneous.

Throughout this section a,b,c are constants.

We study homogeneous equations.

The main tool is characteristic equation for the ODE.

1 distinct real roots

2 repeated real roots

3 complex pair

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Characteristic Equations

Can you guess an exponential solution of 2y⁰⁰+ 5y⁰+ 2y = 0?

Try e^rt:

2y⁰⁰+ 5y⁰+ 2y= (2r²+ 5r+ 2)e^rt = 0 ⇒ 2r²+ 5r+ 2 = 0.

Factoring 2r²+ 5r+ 2 = (2r+ 1)(r+ 2), so we obtain e^−t/2, e^−2t.

Definition

For a second order ODE ay⁰⁰+by⁰+cy = 0, the equation ar²+br+c = 0

is called acharacteristic equation for the ODE.

Such a number r is called acharacteristic root for the ODE.

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Real Distinct Characteristic Roots

Theorem

If r solves ar²+br+c = 0then e^rt is a solution to ay⁰⁰+by⁰+cy = 0.

Theorem

If r1,r2 are real distinct roots of the characteristic equation ar²+br+c = 0,

then e^r¹^t,e^r²^t is a fundamental set of solutions for the ODE ay⁰⁰+by⁰+cy = 0.

Thus

y(t) =C₁e^r¹^t+C₂e^r²^t is a general solution for the ODE.

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Real Distinct Characteristic Roots

Proof. Substituting e^rt into the ODEay⁰⁰+by⁰+cy = 0 we get (ar²+br+c)e^rt = 0 ⇒ ar²+br²+c = 0.

r₁,r₂ solve ar²+br+c = 0, soe^r¹^t,e^r²^t are two solutions of ODE ay⁰⁰+by⁰+cy = 0.

Next, we calculate the Wronskian w(e^r¹^t,e^r²^t) = det

e^r¹^t e^r²^t r1e^r¹^t r2e^r²^t

= (r₂−r₁)e^(r¹^+r²^)t. Since r₂−r₁ 6= 0, we find thatW(e^r¹^t,e^r²^t)6= 0 for all t. So

y(t) =C₁e^r¹^t+C₂e^r²^t is a general solution to the ODE.

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Real Distinct Characteristic Roots

Formula. The quadratic equation

ar²+br+c = 0 has two roots r1,r2 (can be repeated, or complex).

They are given by the formula

r₁ = −b−√

b²−4ac

2a , r₂ = −b+√

b²−4ac

2a .

1 r1,r2 are distinct real numbers ifb²−4ac >0.

2 r1,r2 are repeated (or equal) ifb²−4ac = 0

3 r₁,r₂ are complex pair if b²−4ac <0.

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Real Distinct Characteristic Roots

EX.Find a fundamental set of solutions to the ODE y⁰⁰+ 4y⁰+ 3y = 0.

Then find a general solution formula.

Sol.

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Real Distinct Characteristic Roots

EX.Find a fundamental set of solutions to the ODE 3y⁰⁰−5y⁰−2y= 0.

Sol.

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Real Distinct Characteristic Roots

EX.Find a fundamental set of solutions to the ODE y⁰⁰+ 3y⁰+y = 0.

Sol.

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Exercise

EX.Solve the IVP

y⁰⁰−3y⁰ = 0, y(0) =−1, y⁰(0) = 2.

Sol.

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Abel’s Theorem

We will need the following important result.

Theorem (Abel’s theorem) If y1,y2 are solutions to the ODE

y⁰⁰+p(t)y⁰+q(t)y= 0 then there is a constant C depending on y₁,y₂ such that

W(y₁,y₂)(t) =Ce⁻^R^p(t)dt.

Moreover, if y1,y2 are fundamental set of solutions then C 6= 0.

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Abel’s Theorem

Proof. Since y₁,y₂ are solutions to ODE:

y₁⁰⁰+p(t)y₁⁰ +q(t)y₁= 0 ⇒ y₁⁰⁰ =−p(t)y₁⁰ −q(t)y₁, y₂⁰⁰+p(t)y₂⁰ +q(t)y₂= 0 ⇒ y₂⁰⁰ =−p(t)y₂⁰ −q(t)y₂, . We differentiate the Wronskian W(t) =W(y₁,y₂)(t):

d

dtW = d

dt(y₁y₂⁰ −y₂y₁⁰) =y₁y₂⁰⁰−y₂y₁⁰⁰

=y₁(−p(t)y₂⁰ −q(t)y₂)−y₂(−p(t)y₁⁰ −q(t)y₁)

=−p(t)W.

This is a linear first order ODE: W⁰+p(t)W = 0! So W(t) = 1

I(t)( Z

I(t)(0)dt+C) =Ce⁻

Rp(t)dt.

Ify1,y2 are fundamental solutions W 6= 0, soC 6= 0.

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Abel’s Theorem

EX.Consider ODEy⁰⁰+ 2y⁰+y = 0. The characteristic equation is r²+ 2r+ 1 = 0 ⇒ (r+ 1)²= 0.

So we get−1,−1, which gives only one solutione^−t! Let y1 =e^−t, which is a solution. To findy2, we solve

W(e^−t,y₂) =e⁻^R^2dt. That is

det

e^−t y2

−e^−t y₂⁰

=e^−2t ⇒ y₂⁰ +y₂ =e^−t. Integrating factor I(t) =e^t, so y2 = _e¹t(R

e^te^−tdt+C) =te^−t+Ce^−t. We can choose C = 0 to gety₂ =te^−t!

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Reduction of Order Formula

Theorem (Reduction of order formula)

If y1 6= 0is a solution of ODE y⁰⁰+p(t)y⁰+q(t)y = 0, then y2(t) =y1(t)

Z 1 (y₁(t))²e⁻

Rp(t)dtdt

is a solution of the ODE. Furthermore, {y₁,y₂} is a fundamental set of solutions, hence

y(t) =C₁y₁(t) +C₂y₂(t) is a general solution to the ODE.

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Reduction of Order Formula

Proof. To find y₂, we use Abel’s theorem and solve W(y1,y2) =e⁻

Rp(t)dt.

By the definition of Wronskian, we get

y₁y₂⁰ −y₂y₁⁰ =e⁻^R^p(t)dt ⇒ y₂⁰ −(y₁⁰/y₁)y₂ =e⁻^R^p(t)dt/y₁. This is a linear first order ODE: I(t) =e^R^(−y¹⁰^/y¹^)dt = 1/y₁ and

y2 = 1 I(t)(

Z

I(t)e⁻^R^p(t)dt

y₁ +C) =y1(

Z e⁻^R^p(t)dt

y₁² dt+C).

Setting C = 0, we get a solutiony₂=y₁R _e⁻

Rp(t)dt

y₁² dt.

y2 is chosen so that W(y1,y2) =e⁻^R^p(t)dt 6= 0, so it is automatically that {y₁,y₂} is a fundamental set of solutions.

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Repeated Real Roots

Theorem

If the characteristic equation of ODE ay⁰⁰+by⁰+cy = 0has only one root r (this happens when b²−4ac = 0), then

e^rt, te^rt

form a fundamental set of solutions to the ODE. So y(t) =C1e^rt+C2te^rt is a general solution of the ODE.

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Repeated Real Roots

Proof. Let y₁ =e^rt. By assumptionr solves the characteristic equation, so e^rt is a solution to the ODE.

We use the reduction of order formula:

y2 =y1

Z e⁻^R^(b/a)dt y₁² dt

=e^rt

Z e^−(b/a)t e^2rt dt.

Since ar²+br+c = 0 has a repeated rootr, we getb² = 4ac and r =−b/(2a). So −(b/a) = 2r and hence

y₂=e^rt Z

1dt =te^rt.

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Repeated Real Roots

Sol.

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Repeated Real Roots

EX.Find a fundamental set of solutions to the ODE y⁰⁰+y⁰+¹₄y = 0.

Sol.

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Repeated Real Roots

Solve the IVP

y⁰⁰+ 6y⁰+ 9y = 0, y(0) = 0, y⁰(0) = 2.

Sol.

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Exercise

EX.Determine a general solution to the ODE y⁰⁰−4y⁰+ 4y = 0.

Sol.

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Complex Pair

Now we consider ODE ay⁰⁰+by⁰+cy = 0 that has two complex roots.

This occurs whenb²−4ac <0. Actually, the roots are λ−iµ, λ+iµ,

where λ, µare real numbers andµ6= 0.

Theorem

Ifλ−iµ, λ+iµ(µ6= 0) are roots of the characteristic equation for ODE ay⁰⁰+by⁰+cy = 0, then

e^λtcosµt, e^λtsinµt

form a fundamental set of solutions. So the general solution to the ODE is y(t) =e^λt(C1cosµt+C2sinµt).

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Complex Pair

Proof. Note thatλ=−b/(2a) andµ=√

4ac−b²/(2a). Let y₁ =e^λtcosµt and y₂ =e^λtsinµt. We calculate

y₁⁰ =e^λt(λcosµt−µsinµt),

y₁⁰⁰ =e^λt(λ²cosµt−2λµsinµt−µ²cosµt) So

ay₁⁰⁰+by₁⁰ +cy₁ =e^λt(a(λ²−µ²) +bλ+c) cosµt +e^λt(−2aλµ−µb) sinµt Now −2aλµ−µb= 0 and

a(λ²−µ²) +bλ+c = 2b²−4ac 4a −b²

2a+c = 0.

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Complex Pair

So y₁ solves the ODE. Similarly, y₂ also solves the ODE.

Next, we calculate the Wronskian

W(y1,y2) = det

e^λtcosµt e^λtsinµt e^λt(λcosµt−µsinµt) e^λt(λsinµt+µcosµt)

=e^2λtµ(cos²µt+ sin²µt)

=e^2λtµ.

We have used the trig. identity: cos²A+ sin²A= 1. Since µ6= 0, it follows that the Wronskian is non-zero.

Thus y₁,y₂ is a fundamental set of solutions.

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Complex Pair

EX.Consider ODEy⁰⁰+ 2y⁰+ 2y = 0. Find the characteristic roots and a general solution.

Sol.

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Complex Pair

EX.Find a general solution of the ODE y⁰⁰+ 4y⁰+ 8y= 0.

Sol.

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Exercise

EX.Determine a general solution to the ODE y⁰⁰+ 2y⁰+ 5y = 0.

Sol.

Linear Algebra and Differential Equations