Differential equations – university Year 1 maths lesson to download as PDF

Example 1

Solve \(\dfrac{dy}{dx}=xy\), and solve \(\dfrac{dy}{dx}=y^2\) with \(y(0)=1\).

First: \(\int\frac{dy}y=\int x\,dx\), so \(\ln|y|=\frac{x^2}2+c\) and \(y=Ce^{x^2/2}\). Second: \(\int y^{-2}dy=\int dx\) gives \(-\frac1y=x+c\); \(y(0)=1\) gives \(c=-1\), so \(y=\frac1{1-x}\), which blows up at \(x=1\).

Example 2

Solve \(y'+2y=e^{-x}\), and solve \(y'-y=x\) with \(y(0)=0\).

First: \(\mu=e^{2x}\), so \((e^{2x}y)'=e^{x}\), \(e^{2x}y=e^x+C\) and \(y=e^{-x}+Ce^{-2x}\). Second: \(\mu=e^{-x}\), \((e^{-x}y)'=xe^{-x}\), giving \(y=-x-1+Ce^x\); \(y(0)=0\) gives \(C=1\).

Example 3

Solve \(y''-5y'+6y=0\), \(y''+4y=0\), \(y''+2y'+y=0\) and \(y''+2y'+5y=0\).

The roots are \(2,3\) so \(y=Ae^{2x}+Be^{3x}\); then \(\pm2i\) so \(y=A\cos2x+B\sin2x\); then \(-1\) twice so \(y=(A+Bx)e^{-x}\); then \(-1\pm2i\) so \(y=e^{-x}(A\cos2x+B\sin2x)\), a decaying oscillation.

Example 4: IVP

Solve \(y''-y=0\) with \(y(0)=1\) and \(y'(0)=0\).

The general solution is \(Ae^x+Be^{-x}\). The conditions give \(A+B=1\) and \(A-B=0\), so \(A=B=\frac12\) and \(y=\cosh x\).

Example 5

Solve \(y''-3y'+2y=4\), and find a particular integral for \(y''+y'-2y=e^{3x}\).

First: try \(y_p=2\), and \(y_c=Ae^x+Be^{2x}\), so \(y=Ae^x+Be^{2x}+2\). Second: try \(ke^{3x}\): \((9+3-2)k=1\), so \(k=\frac1{10}\).

Example 6

A drink at \(90^\circ\)C cools in a \(20^\circ\)C room and reaches \(50^\circ\)C after 10 minutes. Find \(k\).

\(T=20+70e^{-kt}\). At \(t=10\): \(\frac{30}{70}=e^{-10k}\), so \(k=\frac{\ln(7/3)}{10}\approx0.0847\).

Worked example 7: A decay problem with a half-life

A radioactive sample satisfies \(\dfrac{dN}{dt}=-\lambda N\) and halves in 8 days. Find \(\lambda\) and the fraction left after 20 days.

Solving gives \(N=N_0e^{-\lambda t}\). The half-life condition \(\tfrac12=e^{-8\lambda}\) gives \(\lambda=\dfrac{\ln2}8\approx0.0866\) per day. After 20 days the fraction left is \(e^{-20\lambda}=2^{-2.5}\approx0.177\), so about 17.7 per cent of the sample remains. Notice that no value of \(N_0\) was needed: the fraction depends only on the rate.

Worked example 8: A separable equation that cannot be solved for y

Solve \(\dfrac{dy}{dx}=\dfrac{x}{y^2}\) with \(y(0)=1\).

Separate to get \(\int y^2dy=\int x\,dx\), so \(\dfrac{y^3}3=\dfrac{x^2}2+c\). The condition gives \(c=\tfrac13\). Multiplying by 3, \(y^3=\tfrac32x^2+1\), and taking a cube root gives \(y=\left(1+\tfrac32x^2\right)^{1/3}\). In other cases an implicit relation between \(x\) and \(y\) is the best you can do, and that is a perfectly acceptable answer.

Worked example 9: An integrating factor with a trigonometric term

Solve \(y'+y\cot x=\cos x\) for \(0\lt x\lt \pi\).

Here \(P=\cot x\), so \(\mu=e^{\int\cot x\,dx}=e^{\ln\sin x}=\sin x\). Then \((y\sin x)'=\sin x\cos x\), and integrating gives \(y\sin x=\tfrac12\sin^2x+C\). So \(y=\tfrac12\sin x+C\csc x\). The integrating factor turned a hard-looking equation into a product-rule problem, which is exactly what it is designed to do.

Worked example 10: Repeated roots

Find the general solution of \(y''-6y'+9y=0\) and the solution with \(y(0)=1\), \(y'(0)=0\).

The auxiliary equation is \((m-3)^2=0\), so \(y=(A+Bx)e^{3x}\). The first condition gives \(A=1\). Differentiating, \(y'=(B+3A+3Bx)e^{3x}\), so \(y'(0)=B+3=0\) and \(B=-3\). The solution is \(y=(1-3x)e^{3x}\). It crosses zero at \(x=\tfrac13\) and then becomes large and negative.

Worked example 11: A particular integral for a polynomial right-hand side

Find a particular integral for \(y''+3y'+2y=4x+2\).

The right-hand side is a polynomial of degree 1, so try \(y_p=ax+b\). Then \(y_p'=a\) and \(y_p''=0\), so the left side is \(3a+2ax+2b\). Comparing the coefficients of \(x\) gives \(2a=4\), so \(a=2\). Comparing constants gives \(3a+2b=2\), so \(b=-2\). A particular integral is \(y_p=2x-2\). Neither \(1\) nor \(x\) appears in the complementary function \(Ae^{-x}+Be^{-2x}\), so no extra factor of \(x\) was needed.

Worked example 12: The logistic equation

A population satisfies \(y'=y(1-y)\) with \(y(0)=\tfrac12\). Solve it.

Partial fractions give \(\dfrac1{y(1-y)}=\dfrac1y+\dfrac1{1-y}\), so \(\ln\left|\dfrac y{1-y}\right|=x+c\). With \(y(0)=\tfrac12\) the left side is \(\ln1=0\), so \(c=0\) and \(\dfrac y{1-y}=e^x\). Solving for \(y\) gives \(y=\dfrac{e^x}{1+e^x}=\dfrac1{1+e^{-x}}\). The graph is an S-shaped curve that rises from near 0 to near 1.

Worked example 13: A mixing problem with no inflow of salt

A tank holds 100 litres of brine with 30 kg of salt. Pure water flows in at 4 litres per minute and the mixture leaves at the same rate. Find the salt left after 25 minutes.

Salt only leaves, at rate \(4\times\dfrac Q{100}=\dfrac Q{25}\) kg per minute, so \(\dfrac{dQ}{dt}=-\dfrac Q{25}\). This is exponential decay, with solution \(Q=30e^{-t/25}\). After 25 minutes \(Q=30e^{-1}\approx11.0\) kg, so about 37 per cent of the salt remains. After a further 25 minutes it would be multiplied by \(e^{-1}\) again: constant proportional decay is the signature of this equation.

Worked example 14: Finding the steady state first

For \(\dfrac{dy}{dt}=6-2y\) with \(y(0)=10\), find the steady state and then the full solution.

The steady state is where the derivative is zero: \(6-2y=0\), so \(y=3\). Writing \(z=y-3\) turns the equation into \(z'=-2z\), so \(z=z_0e^{-2t}\) with \(z_0=10-3=7\). Therefore \(y=3+7e^{-2t}\). The solution approaches 3 from above, and the gap \(y-3\) halves roughly every \(\ln2/2\approx0.35\) time units.

Worked example 15: Underdamped oscillations

Describe the long-term behaviour of \(y''+2y'+26y=0\).

The auxiliary equation has roots \(-1\pm5i\), so \(y=e^{-t}(A\cos5t+B\sin5t)\). The oscillation has angular frequency 5, so its period is \(\dfrac{2\pi}5\approx1.26\), and the factor \(e^{-t}\) makes the amplitude shrink to about 37 per cent each time unit. The solution therefore tends to zero while changing sign again and again: a decaying oscillation, like a plucked guitar string.