Differential equations – university Year 1 maths test answers to download as PDF

← Back to the test

1. Answer to Question 1 (4 marks)

   The equation is separable, so collect the \(y\) terms on the left and the \(x\) terms on the right: \(\int y\,dy=\int x\,dx\) (1). Integrating both sides gives \(\dfrac{y^2}2=\dfrac{x^2}2+c\) (1). Using the initial condition \(y(0)=2\): \(2=0+c\), so \(c=2\) and \(y^2=x^2+4\) (1). Since we need \(y>0\), we take the positive square root: \(y=\sqrt{x^2+4}\) (1).

   Check: differentiating \(y^2=x^2+4\) implicitly gives \(2yy'=2x\), so \(y'=\tfrac xy\), and \(y(0)=\sqrt4=2\). The negative root is rejected because it fails the initial condition. The solution curve is the upper branch of the hyperbola \(y^2-x^2=4\).

2. Answer to Question 2 (4 marks)

   The equation is linear with \(P=1\), so the integrating factor is \(\mu=e^{\int1\,dx}=e^x\). Multiplying through gives \((e^xy)'=2e^{2x}\) (1). Integrating both sides, \(e^xy=e^{2x}+C\) (1), so the general solution is \(y=e^x+Ce^{-x}\). The condition \(y(0)=2\) gives \(1+C=2\), so \(C=1\) (1). Therefore \(y=e^x+e^{-x}=2\cosh x\) (1). Check: \(y'+y=(e^x-e^{-x})+(e^x+e^{-x})=2e^x\), as required.

   A common error is to forget to multiply the right-hand side by the integrating factor, which would give \((e^xy)'=2e^x\) and the wrong solution \(y=2+Ce^{-x}\). The constant \(C\) must be added before dividing by \(e^x\).

3. Answer to Question 3 (4 marks)

   The auxiliary equation is \(m^2-4m+4=(m-2)^2=0\), which has the repeated root \(2\), so the general solution is \(y=(A+Bx)e^{2x}\) (1). Putting \(x=0\) gives \(y(0)=A=1\) (1). Differentiating with the product rule, \(y'=(B+2A+2Bx)e^{2x}\), so \(y'(0)=B+2A=B+2=3\), which gives \(B=1\) (1). Hence the solution is \(y=(1+x)e^{2x}\) (1). Check: \(y(0)=1\) and \(y'=(3+2x)e^{2x}\) gives \(y'(0)=3\), so both conditions hold.

   The repeated root is the reason for the factor \(x\): writing \(y=Ae^{2x}+Be^{2x}\) gives only one independent solution and cannot satisfy both conditions.

4. Answer to Question 4 (4 marks)

   1. Let \(u=T-20\), the excess temperature. Then \(u'=-ku\), so \(u=u_0e^{-kt}\) with \(u_0=200-20=180\), which gives \(T=20+180e^{-kt}\) (1). At \(t=10\): \(110=20+180e^{-10k}\), so \(e^{-10k}=\dfrac12\) (1). Taking logarithms, \(k=\dfrac{\ln2}{10}\approx0.0693\) per minute (1).
   2. Set \(50=20+180e^{-kt}\), so \(e^{-kt}=\dfrac16\) and \(t=\dfrac{\ln6}{k}=\dfrac{10\ln6}{\ln2}\approx\) 25.8 minutes (1). The bar halves its excess temperature every 10 minutes, so this answer is sensible: it lies between 20 minutes (excess 45) and 30 minutes (excess 22.5).

5. Answer to Question 5 (4 marks)

   First solve the homogeneous equation. The auxiliary equation is \(m^2+m-2=(m+2)(m-1)=0\), with roots \(1\) and \(-2\), so the complementary function is \(y_c=Ae^x+Be^{-2x}\) (2 marks). Since \(2\) is not a root, try a particular integral \(y_p=ke^{2x}\). Then \(y_p'=2ke^{2x}\), \(y_p''=4ke^{2x}\), and substituting gives \((4+2-2)k=6\), so \(k=\tfrac32\) (1). The general solution is the sum: \(y=Ae^x+Be^{-2x}+\tfrac32e^{2x}\) (1).

   Check the particular integral: \(\tfrac32(4+2-2)e^{2x}=6e^{2x}\). If the right-hand side had been \(e^x\), the trial function would need to be \(kxe^x\) because 1 is a root of the auxiliary equation.

Total : 20 marks