Differential equations (Further Pure 2)
Introduced| English | Meaning |
|---|---|
| differential equation/ˌdɪfəˈrenʃl ɪˈkweɪʒn/ | An equation containing a derivative of an unknown function, such as velocity and its rate of change. |
| function/ˈfʌŋkʃn/ | A mapping that assigns exactly one output to each allowed input. |
| first-order linear equation/ˌfɜːst ˈɔːdə ˈlɪniə ɪˈkweɪʒn/ | An equation of the form y prime plus P(x) times y equals Q(x), on an interval where its coefficients are defined. |
| integrating factor/ˈɪntɪɡreɪtɪŋ ˈfæktə/ | A factor used to multiply a differential equation so that an expression becomes a complete derivative and can be integrated. |
| general solution/ˈdʒenərəl səˈluːʃn/ | A family of solutions containing arbitrary constants or parameters that represents all solutions of the specified type. |
| initial condition/ɪˈnɪʃl kənˈdɪʃn/ | A given value of the solution at its starting time or starting state. |
| complementary function/ˌkɒmplɪˈmentəri ˈfʌŋkʃn/ | The general solution of the corresponding homogeneous linear differential equation. |
| particular integral/pəˈtɪkjʊlə ˈɪntɪɡrəl/ | Any one solution of the full non-homogeneous linear differential equation; it contains no arbitrary constants. |
| particular solution/pəˈtɪkjʊlə səˈluːʃn/ | A complete solution whose arbitrary constants have been fixed by conditions. |
| auxiliary equation/ɔːkˈsɪlɪəri ɪˈkweɪʒn/ | For a linear differential equation with constant coefficients, the polynomial equation obtained by substituting an exponential trial solution. |
| resonance/ˈrezənəns/ | In the trial method, overlap with the complementary function requires extra factors of x. An undamped system forced at its natural frequency can have a growing response. |
| boundary condition/ˈbaʊndəri kənˈdɪʃn/ | A condition on a solution or its derivatives at a specified point or endpoint, used to determine constants. |
| transient/ˈtrænsɪənt/ | A contribution to a solution that tends to zero as the independent variable increases. |
| amplitude/ˈæmplɪtjuːd/ | The maximum displacement from an equilibrium position. |
| phase lag/ˈfeɪz læɡ/ | The angle by which an oscillation falls behind a stated reference oscillation. |
| separable variables/ˈsepərəbl ˈveərɪəblz/ | A differential equation that can be rearranged as f(y)dy = g(x)dx for direct integration. |
Balance what enters and leaves
- A tank has 50 litres of well-mixed liquid. Each minute, 10 litres enters carrying 2 grams of salt, and 10 litres leaves. If $M$ is salt mass in grams and $t$ is time in minutes, the outflow removes $10(M/50)=M/5$ grams per minute.
- The differential equation 微分方程 is $M'=2-M/5$. It links a function 函数 to its rate of change. We will find the mass, use the starting mass and check what happens after a long time.
Differential equations
dy/dx = a y
For the separate simple equation y′=0.5y, slope is proportional to y. This widget does not model the tank equation M′=2−M/5.
Put a first-order equation in linear form
- A first-order linear equation 一阶线性方程 has the form $y'+P(x)y=Q(x)$. First divide by the coefficient of $y'$, on an interval where it is non-zero. Here a prime means differentiation with respect to the independent variable.
- For $2xy'-4y=2x^3$, division gives $y'-2y/x=x^2$, so $P=-2/x$, not $-4$. Use an interval such as $x>0$; do not extend the result across $x=0$ without checking the original equation.
Build the integrating factor from the product rule
- We want $\mu y'+\mu Py=(\mu y)'$. The product rule requires $\mu'=P\mu$, giving an integrating factor 积分因子 $\mu=e^{\int P\,dx}$. Any non-zero constant multiple works; choose the simplest one.
- Multiply every term by $\mu$, integrate, then divide by $\mu$. Keep one arbitrary constant in the general solution 通解:
For dy/dx + 2y = eˣ, the integrating factor μ = e^(∫2 dx) is:
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∫2 dx = 2x, so μ = e^(2x).
Multiplying by the integrating factor makes the left side d/dx(μy), so you can integrate directly.
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That is the whole point of the integrating factor — it makes the left side an exact derivative.
Use the product rule to explain why $\mu=e^{\int P\,dx}$ works for $y′+Py=Q$. Does multiplying μ by 7 change the general solution?
The factor is designed to create one product derivative.
Finish an algebraic integrating-factor example
- For $y'-2y/x=x^2$ on $x>0$, choose $\mu=e^{-2\ln x}=x^{-2}$. Multiplying gives $(x^{-2}y)'=1$, not $x^2$.
- Integrate to get $x^{-2}y=x+C$, so $y=x^3+Cx^2$. If $y(1)=2$, then $C=1$ and $y=x^3+x^2$. Substitution gives $2xy'-4y=2x^3$.
Solve $3xy′+6y=3x^3$ for $x>0$, given $y(1)=1$. Show the normalised equation and integrating factor.
Normalising fixes P and Q before the integrating factor is chosen.
Let the integrating factor reveal a standard integral
- Consider $y'-\dfrac{x+2}{(x+2)^2+9}y=1$, with $y(1)=0$. The numerator is half the derivative of the denominator. Thus $\mu=((x+2)^2+9)^{-1/2}$ and
- Use the inverse-hyperbolic integral from the previous topic. The initial condition 初始条件 fixes the constant:
Here $\sinh^{-1}$ means inverse hyperbolic sine, not a reciprocal.
Solve $y′-\dfrac{x+1}{(x+1)^2+16}y=1$ with $y(3)=0$. Give an exact answer.
A half-log factor changes the remaining integral into an inverse-hyperbolic form.
Use hyperbolic factors on a stated interval
- For $y'+y\coth x=\cosh x$ on $x>0$, $\int\coth x\,dx=\ln(\sinh x)$. Choose $\mu=\sinh x$; then $(y\sinh x)'=\sinh x\cosh x$.
- Hence $y\sinh x=\tfrac12\sinh^2x+C$ and $y=\tfrac12\sinh x+C\operatorname{csch}x$. If $y(\ln2)=1$, use $\sinh(\ln2)=3/4$ to find $C=15/32$. The equation itself is undefined at $x=0$.
Solve $y′+y\tanh x=\sinh x$ with $y(0)=2$.
A hyperbolic integrating factor can turn the right side into a direct substitution integral.
Separate the complementary and particular parts
- For a linear equation $L[y]=f(x)$, the complementary function 余函数 (CF) solves $L[y]=0$. A particular integral 特积分 (PI) is any one solution of $L[y]=f(x)$. Linearity gives $y=\mathrm{CF}+\mathrm{PI}$.
- This also works at first order: for $y'+2y=e^x$, CF $=Ce^{-2x}$ and PI $=e^x/3$. A particular solution 特解 is the complete solution after its arbitrary constants have been fixed; it is not another name for the PI.
For a constant-coefficient linear equation, the general solution is:
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General solution = complementary function (RHS = 0) + a particular integral.
For $y′+2y=e^x$, explain why $e^x/3$ is not the general solution. If $y(0)=1$, distinguish PI from particular solution.
A PI is one full-equation solution; conditions select one member of CF plus PI.
Find the CF for $y′-3y=4x$. Then find a PI by trying $ax+b$.
The CF/PI method also applies to a first-order constant-coefficient equation.
Find all three complementary-function forms
- For $ay''+by'+cy=0$ with constant coefficients and $a\ne0$, substituting $y=e^{mx}$ gives the auxiliary equation 辅助方程 $am^2+bm+c=0$. Use two independent terms, including the extra $x$ for a repeated root.
- These examples cover distinct, repeated and conjugate-complex roots:
| equation | roots | CF |
|---|---|---|
| $y''-3y'+2y=0$ | $1,2$ | $Ae^x+Be^{2x}$ |
| $y''+4y'+4y=0$ | $-2,-2$ | $(A+Bx)e^{-2x}$ |
| $y''+2y'+5y=0$ | $-1\pm2i$ | $e^{-x}(A\cos2x+B\sin2x)$ |
For y″ − 5y′ + 6y = 0 the auxiliary equation is m² − 5m + 6 = 0. What is its larger root?
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m² − 5m + 6 = (m − 2)(m − 3) = 0, so m = 2 or 3; the larger is 3.
Match each type of auxiliary root to the complementary function it gives.
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Distinct reals give two exponentials; a repeated root adds a factor of x; complex roots give an oscillating, exponentially-scaled form.
For y double prime minus 7 y prime plus 10 y = 0, what is the larger root of the auxiliary equation, and what changes if a root is repeated?
The auxiliary polynomial is built from the constant-coefficient equation. Distinct real roots each contribute one exponential.
Find the CF for each equation: (i)$y″-5y′+6y=0$; (ii)$y″-6y′+9y=0$; (iii)$y″+4y′+13y=0$.
The real and imaginary parts of auxiliary roots control exponential scale and frequency.
Match every power in a polynomial particular integral
- For $y''+y'-2y=2x^2+3x+1$, try $y_p=ax^2+bx+c$. Include all lower powers even if the right side omits one. Substitute $y_p'=2ax+b$ and $y_p''=2a$.
- Match coefficients, starting with the highest power:
Thus $a=-1$, $b=-5/2$, $c=-11/4$ and $y_p=-x^2-5x/2-11/4$.
Find a PI for $y″-y′-2y=2x^2+3x+1$ by matching coefficients.
Matching every power finds all coefficients of the polynomial PI.
Try an exponential and substitute its derivatives
- For $y''+2y'+5y=4e^{-x}$, try $y_p=ke^{-x}$. Its derivatives are $-ke^{-x}$ and $ke^{-x}$, so $(1-2+5)ke^{-x}=4e^{-x}$ and $k=1$.
- Combine it with the complex-root CF. If $y(0)=y'(0)=0$, the full solution is $y=e^{-x}(1-\cos2x)$. The derivative condition must include the PI derivative as well as the CF derivative.
Solve $y″+4y′+8y=5e^{-2x}$ with $y(0)=y′(0)=0$.
The initial derivative must include both the exponential factor and the oscillation.
Keep both sine and cosine in the trial
- For $y''+2y'+2y=2\sin2x+6\cos2x$, try $y_p=a\sin2x+b\cos2x$. Differentiation mixes the two functions, even if the forcing has only one of them.
- Substitution and coefficient matching give
So $y_p=\sin2x-\cos2x$ and the full solution adds $e^{-x}(A\cos x+B\sin x)$.
Find the general solution of $y″+2y′+2y=6\sin2x-2\cos2x$.
Differentiation mixes sine and cosine, so the trial needs both.
Repair a trial that is already in the CF
- If a trial is in the CF, applying $L$ gives zero, so no coefficient can match a non-zero right side. Multiply the trial by $x$ until it is independent of the CF. This is the resonance 共振 adjustment in the trial method.
- For $y''+4y=\sin2x$, $\sin2x$ is in the CF. Use $y_p=kx\cos2x$; substitution gives $y_p''+4y_p=-4k\sin2x$, so $k=-1/4$. Thus PI $=-x\cos2x/4$.
Solve $y′-3y=2e^{3x}$. Explain why the trial $ke^{3x}$ cannot work.
A homogeneous term cannot produce non-zero forcing without an extra factor of x.
Find a PI for $y″+9y=\cos3x$. Verify its coefficient by substitution.
The sign of the resonant coefficient depends on the supplied forcing.
Count repeated overlaps before choosing the power of x
- For $y''-4y'+4y=e^{2x}$, the auxiliary root $2$ is repeated. Both $e^{2x}$ and $xe^{2x}$ already occur in the CF, so the trial is $y_p=kx^2e^{2x}$.
- Differentiating and substituting gives $y_p''-4y_p'+4y_p=2ke^{2x}$. Hence $k=1/2$ and
Find the general solution of $y″+2y′+y=6e^{-x}$. Why is a single factor x insufficient?
The multiplicity of an overlapping root determines how many factors of x are needed.
Use a supplied form even when it is unfamiliar
- A question may supply a PI form outside the standard polynomial/exponential/sine-cosine list. Differentiate that form, substitute into the equation and solve for its coefficients; do not guess an extra family.
- Given $y_p=k/x$ for $y''+y=2/x^3+1/x$ on $x>0$, $y_p''=2k/x^3$. Substitution gives $2k/x^3+k/x=2/x^3+1/x$, so $k=1$. The full solution is $A\cos x+B\sin x+1/x$.
You are told that $kx\cos x$ is a PI of $y″+y=-4\sin x$. Find k and give the general solution.
A supplied PI form is checked directly by differentiation and substitution.
Apply conditions to the complete solution
- Return to $y''+y'-2y=2x^2+3x+1$. Its complete solution is $Ae^x+Be^{-2x}-x^2-5x/2-11/4$. Differentiate before applying $y(0)=y'(0)=0$.
- The conditions give $A+B=11/4$ and $A-2B=5/2$, so $A=8/3$, $B=1/12$. Two independent conditions usually determine two constants; some boundary conditions 边界条件 can instead be inconsistent or leave a constant free.
The arbitrary constants A and B are fixed by the ______ conditions.
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Boundary or initial conditions (like y(0) = 1) determine A and B.
For $y″-y′-2y=2x^2+3x+1$, use your PI to find the solution with $y(0)=y′(0)=0$.
Initial conditions apply after adding the CF and PI.
For $y″+y=0$, can $y(0)=0$ and $y(\pi)=1$ be satisfied? What if the second condition is $y(\pi)=0$?
Two boundary conditions can be dependent or inconsistent.
Interpret the transient and the lasting response
- For $y''+2y'+2y=2\sin2x+6\cos2x$, the transient 暂态 term is $e^{-x}(A\cos x+B\sin x)$. It tends to zero as $x\to\infty$, whatever the finite starting constants.
- The lasting response is $y\approx\sin2x-\cos2x=\sqrt2\sin(2x-\pi/4)$. The amplitude 振幅 is $\sqrt2$, the angular frequency is $2$ and the phase lag 相位滞后 relative to $\sin2x$ is $\pi/4$. A positive auxiliary-root real part would prevent this decay argument.
For $y″+2y′+2y=6\sin2x-2\cos2x$, express the large-positive-x response as $R\sin(2x-\phi)$ with $R>0$ and $0<\phi<\pi$.
Amplitude and phase follow coefficient matching with a justified decay limit.
Change independent variable using the chain rule
- If $x=e^t$, write $Y(t)=y(e^t)$ and $t=\ln x$, so $x>0$. The chain rule gives $Y_t=xy_x$; differentiating once more gives $Y_{tt}=xy_x+x^2y_{xx}$.
- Therefore the correct replacement is $x^2y_{xx}=Y_{tt}-Y_t$, not just $Y_{tt}$:
With $x=e^t$ and $Y(t)=y(e^t)$, derive the replacements for $xy′$ and $x^2y″$. Transform $2x^2y″+3xy′-y=x$.
Changing the independent variable requires two applications of the chain rule.
Solve the transformed equation and change back
- Use $x=e^t$ in $x^2y''-3xy'+4y=x^2$. The transformed equation is $Y_{tt}-4Y_t+4Y=e^{2t}$, with the repeated-resonance solution just found.
- Change back to the original variable before applying the stated conditions. Since $y(1)=0$ and $y'(1)=1$, $A=0$, $B=1$:
Use $x=e^t$ to solve $x^2y″+xy′-4y=x^3$ on $x>0$, with $y(1)=1$, $y′(1)=0$.
The original derivative condition must be applied in the original variable.
Use $x=e^t$ to find the general solution of $xy′+2y=x^3$ on $x>0$.
The exponential substitution can reduce a first-order equation too.
Use y equals ux to separate variables
- Given $xy'=y+\sqrt{4x^2+y^2}$ and $y(1)=0$, work on $x>0$ and set $y=ux$. The product rule gives $y'=u+xu'$. Also $\sqrt{x^2(4+u^2)}=x\sqrt{4+u^2}$ on this interval.
- After cancelling $xu$, use separable variables 可分离变量 and integrate:
The condition gives $C=0$, so $u=2\sinh(\ln x)=x-1/x$ and $y=x^2-1$. Check the positive square-root sign in the original equation.
Use $y=ux$ to solve $xy′=y+\sqrt{16x^2+y^2}$ with $y(1)=0$, on $x>0$.
The sign of x matters when a factor is removed from a square root.
Use $y=ux$ to solve $y′=(x+y)/(x-y)$ near $x=1$, given $y(1)=0$. An implicit answer is acceptable.
A ratio substitution separates the equation after the product rule.
A sum substitution can expose a lost solution
- Given $y'=(1+2x+2y)/(2x+2y-1)$, put $z=x+y$. Then $z'=1+y'=4z/(2z-1)$. For $z\ne0$, integrate $(1/2-1/(4z))\,dz=dx$.
- With $y(1)=0$, the local branch has $z>1/2$ and the result is
Dividing by $z$ loses the separate solution $z=0$, or $y=-x$, which does not meet this initial condition. The original equation is undefined at $z=1/2$.
Put $z=x+y$ in $y′=(1+4x+4y)/(4x+4y-1)$. Find an implicit solution through $(1,0)$ and identify any solution lost by dividing.
A sum substitution changes the derivative by one; division can lose a constant transformed solution.
Given $y′=y+e^x/y$ and $y(0)=1$, use $u=y^2$ to find the positive solution and its interval.
A square substitution requires a branch choice when returning to y.
Answer the model question in its own quantities
- For the tank equation $M'+M/5=2$, an integrating factor is $e^{t/5}$, giving $M=10+Ce^{-t/5}$. Initially fresh liquid means $M(0)=0$, so $M=10(1-e^{-t/5})$ grams.
- Half the limiting mass is reached at $t=5\ln2$ minutes. As $t\to\infty$, $M\to10$ grams and inflow/outflow salt rates both approach 2 grams per minute. The model assumes constant volume, fixed input and immediate uniform mixing.
A well-mixed tank contains 40 litres. Each minute 10 litres enters with 3 grams of salt, and 10 litres leaves. Initially it contains 4 grams. Find M(t), the limiting mass and the time to reach 8 grams.
A salt balance uses the concentration leaving the tank, not the initial concentration.
Choose a method and verify the original equation
- Normalise a first-order linear equation before choosing its integrating factor. For constant coefficients, find CF, choose and check PI, add them, then apply all conditions. For a given substitution, transform derivatives as well as variables.
- Differentiate the final answer and substitute it into the original equation, check the conditions, retain the allowed interval and interpret any units, limits or model assumptions. A correct transformed equation is not yet a solution in the original variable.
For y″ − 7y′ + 10y = 0, what is the larger auxiliary root?
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The roots are 2 and 5, so the larger root is 5. Other options — 2 (the smaller root); 3 (the lesson larger root of m squared minus 5m plus 6).
A student says every non-homogeneous equation can use a polynomial PI, and every CF disappears at large x. Refute both claims using equations from this lesson.
Forcing type and auxiliary roots must both be examined before making a trial or limit.