Basics of Differential Equations

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Verify that the function \(y={e}^{-3x}+2x+3\) is a solution to the differential equation \({y}^{'}+3y=6x+11.\)

   Jiżvelaw it-tweġiba

   To verify the solution, we first calculate \({y}^{'}\) using the chain rule for derivatives. This gives \({y}^{'}=-3{e}^{-3x}+2.\) Next we substitute \(y\) and \({y}^{'}\) into the left-hand side of the differential equation:

   \[(-3{e}^{-3x}+2)+3({e}^{-3x}+2x+3).\]

   The resulting expression can be simplified by first distributing to eliminate the parentheses, giving

   \[-3{e}^{-3x}+2+3{e}^{-3x}+6x+9.\]

   Combining like terms leads to the expression \(6x+11,\) which is equal to the right-hand side of the differential equation. This result verifies that \(y={e}^{-3x}+2x+3\) is a solution of the differential equation.

2. Verify that \(y=2{e}^{3x}-2x-2\) is a solution to the differential equation \({y}^{'}-3y=6x+4.\)

3. What is the order of each of the following differential equations?

   1. \({y}^{'}-4y={x}^{2}-3x+4\)
   2. \({x}^{2}y\text{‴}-3x{y}^{″}+x{y}^{'}-3y=\text{sin}\ x\)
   3. \(\frac{4}{x}{y}^{(4)}-\frac{6}{{x}^{2}}{y}^{″}+\frac{12}{{x}^{4}}y={x}^{3}-3{x}^{2}+4x-12\)
   Jiżvelaw it-tweġiba

   1. The highest derivative in the equation is \({y}^{'},\) so the order is \(1.\)
   2. The highest derivative in the equation is \(y\text{‴}\text{,}\) so the order is \(3.\)
   3. The highest derivative in the equation is \({y}^{(4)},\) so the order is \(4.\)

4. What is the order of the following differential equation?

   \[({x}^{4}-3x){y}^{(5)}-(3{x}^{2}+1){y}^{'}+3y=\text{sin}\ x\ \text{cos}\ x\]
   Jiżvelaw it-tweġiba

   \(5\)

5. Find the particular solution to the differential equation \({y}^{'}=2x\) passing through the point \((2,7).\)

   Jiżvelaw it-tweġiba

   Any function of the form \(y={x}^{2}+C\) is a solution to this differential equation. To determine the value of \(C,\) we substitute the values \(x=2\) and \(y=7\) into this equation and solve for \(C\text{:}\)

   \[\begin{array}{l} \\ \\ y={x}^{2}+C \\ 7={2}^{2}+C=4+C \\ C=3.\end{array}\]

   Therefore the particular solution passing through the point \((2,7)\) is \(y={x}^{2}+3.\)

6. Find the particular solution to the differential equation

   \[{y}^{'}=4x+3\]

   passing through the point \((1,7),\) given that \(y=2{x}^{2}+3x+C\) is a general solution to the differential equation.

   Jiżvelaw it-tweġiba

   \(y=2{x}^{2}+3x+2\)

7. Verify that the function \(y=2{e}^{-2t}+{e}^{t}\) is a solution to the initial-value problem

   \[{y}^{'}+2y=3{e}^{t},\ y(0)=3.\]
   Jiżvelaw it-tweġiba

   For a function to satisfy an initial-value problem, it must satisfy both the differential equation and the initial condition. To show that \(y\) satisfies the differential equation, we start by calculating \({y}^{'}.\) This gives \({y}^{'}=-4{e}^{-2t}+{e}^{t}.\) Next we substitute both \(y\) and \({y}^{'}\) into the left-hand side of the differential equation and simplify:

   \[\begin{array}{ll}{y}^{'}+2y & =(-4{e}^{-2t}+{e}^{t})+2(2{e}^{-2t}+{e}^{t}) \\ & =-4{e}^{-2t}+{e}^{t}+4{e}^{-2t}+2{e}^{t} \\ & =3{e}^{t}.\end{array}\]

   This is equal to the right-hand side of the differential equation, so \(y=2{e}^{-2t}+{e}^{t}\) solves the differential equation. Next we calculate \(y(0)\text{:}\)

   \[\begin{array}{ll}y(0) & =2{e}^{-2(0)}+{e}^{0} \\ & =2+1 \\ & =3.\end{array}\]

   This result verifies the initial value. Therefore the given function satisfies the initial-value problem.

8. Verify that \(y=3{e}^{2t}+4\ \text{sin}\ t\) is a solution to the initial-value problem

   \[{y}^{'}-2y=4\ \text{cos}\ t-8\ \text{sin}\ t,\ y(0)=3.\]

9. Solve the following initial-value problem:

   \[{y}^{'}=3{e}^{x}+{x}^{2}-4,\ y(0)=5.\]
   Jiżvelaw it-tweġiba

   The first step in solving this initial-value problem is to find a general family of solutions. To do this, we find an antiderivative of both sides of the differential equation

   \[\int {y}^{'}dx=\int (3{e}^{x}+{x}^{2}-4)\ dx,\]

   namely,

   \[y+{C}_{1}=3{e}^{x}+\frac{1}{3}{x}^{3}-4x+{C}_{2}.\]

   We are able to integrate both sides because the y term appears by itself. Notice that there are two integration constants: \({C}_{1}\) and \({C}_{2}.\) Solving for \(y\) gives

   \[y=3{e}^{x}+\frac{1}{3}{x}^{3}-4x+{C}_{2}-{C}_{1}.\]

   Because \({C}_{1}\) and \({C}_{2}\) are both constants, \({C}_{2}-{C}_{1}\) is also a constant. We can therefore define \(C={C}_{2}-{C}_{1},\) which leads to the equation

   \[y=3{e}^{x}+\frac{1}{3}{x}^{3}-4x+C.\]

   Next we determine the value of \(C.\) To do this, we substitute \(x=0\) and \(y=5\) into and solve for \(C\text{:}\)

   \[\begin{array}{lll} \\ 5 & = & 3{e}^{0}+\frac{1}{3}{0}^{3}-4(0)+C \\ 5 & = & 3+C \\ C & = & 2.\end{array}\]

   Now we substitute the value \(C=2\) into . The solution to the initial-value problem is \(y=3{e}^{x}+\frac{1}{3}{x}^{3}-4x+2.\)

10. Solve the initial-value problem

    \[{y}^{'}={x}^{2}-4x+3-6{e}^{x},\ y(0)=8.\]
    Jiżvelaw it-tweġiba

    \(y=\frac{1}{3}{x}^{3}-2{x}^{2}+3x-6{e}^{x}+14\)

11. A baseball is thrown upward from a height of \(3\) meters above Earth’s surface with an initial velocity of \(10\ \text{m/s},\) and the only force acting on it is gravity. The ball has a mass of \(0.15\ \text{kg}\).

    1. Find the velocity \(v(t)\) of the baseball at time \(t.\)
    2. What is its velocity after \(2\) seconds?
    Jiżvelaw it-tweġiba

    1. From the preceding discussion, the differential equation that applies in this situation is
       \[{v}^{'}(t)=\text{-}g,\]
       where \(g=9.8{\ \text{m/s}}^{2}.\) The initial condition is \(v(0)={v}_{0},\) where \({v}_{0}=10\ \text{m/s}\text{.}\) Therefore the initial-value problem is \({v}^{'}(t)=-9.8{\ \text{m/s}}^{2},v(0)=10\ \text{m/s}\text{.}\)
       The first step in solving this initial-value problem is to take the antiderivative of both sides of the differential equation. This gives
       \[\begin{array}{lll}\int {v}^{'}(t)\ dt & = & \int -9.8dt \\ v(t) & = & -9.8t+C.\end{array}\]
       The next step is to solve for \(C.\) To do this, substitute \(t=0\) and \(v(0)=10\text{:}\)
       \[\begin{array}{lll}v(t) & = & -9.8t+C \\ v(0) & = & -9.8(0)+C \\ 10 & = & C.\end{array}\]
       Therefore \(C=10\) and the velocity function is given by \(v(t)=-9.8t+10.\)
    2. To find the velocity after \(2\) seconds, substitute \(t=2\) into \(v(t).\)
       \[\begin{array}{lll}v(t) & = & -9.8t+10 \\ v(2) & = & -9.8(2)+10 \\ v(2) & = & -9.6.\end{array}\]
       The units of velocity are meters per second. Since the answer is negative, the object is falling at a speed of \(9.6\ \text{m/s}\text{.}\)

12. Suppose a rock falls from rest from a height of \(100\) meters and the only force acting on it is gravity. Find an equation for the velocity \(v(t)\) as a function of time, measured in meters per second.

    Jiżvelaw it-tweġiba

    \(v(t)=-9.8t\)

13. A baseball is thrown upward from a height of \(3\) meters above Earth’s surface with an initial velocity of \(10\ \text{m/s},\) and the only force acting on it is gravity. The ball has a mass of \(0.15\) kilogram.

    1. Find the position \(s(t)\) of the baseball at time \(t.\)
    2. What is its height after \(2\) seconds?
    Jiżvelaw it-tweġiba

    1. We already know the velocity function for this problem is \(v(t)=-9.8t+10.\) The initial height of the baseball is \(3\) meters, so \({s}_{0}=3.\) Therefore the initial-value problem for this example is
       To solve the initial-value problem, we first find the antiderivatives:
       \[\begin{array}{lll}\int {s}^{'}(t)\ dt & = & \int -9.8t+10dt \\ s(t) & = & -4.9{t}^{2}+10t+C.\end{array}\]
       Next we substitute \(t=0\) and solve for \(C\text{:}\)
       \[\begin{array}{lll}s(t) & = & -4.9{t}^{2}+10t+C \\ s(0) & = & -4.9{(0)}^{2}+10(0)+C \\ 3 & = & C.\end{array}\]
       Therefore the position function is \(s(t)=-4.9{t}^{2}+10t+3.\)
    2. The height of the baseball after \(2\ \text{s}\) is given by \(s(2)\text{:}\)
       \[\begin{array}{ll}s(2) & =-4.9{(2)}^{2}+10(2)+3 \\ & =-4.9(4)+23 \\ & =3.4.\end{array}\]
       Therefore the baseball is \(3.4\) meters above Earth’s surface after \(2\) seconds. It is worth noting that the mass of the ball cancelled out completely in the process of solving the problem.

14. \({y}^{'}+y=3{y}^{2}\)

    Jiżvelaw it-tweġiba

    \(1\)

15. \({({y}^{'})}^{2}={y}^{'}+2y\)

16. \(y\text{‴}+{y}^{″}{y}^{'}=3{x}^{2}\)

    Jiżvelaw it-tweġiba

    \(3\)

17. \({y}^{'}={y}^{″}+3{t}^{2}\)

18. \(\frac{dy}{dt}=t\)

    Jiżvelaw it-tweġiba

    \(1\)

19. \(\frac{dy}{dx}+\frac{{d}^{2}y}{d{x}^{2}}=3{x}^{4}\)

20. \({(\frac{dy}{dt})}^{2}+8\frac{dy}{dt}+3y=4t\)

    Jiżvelaw it-tweġiba

    \(1\)

21. \(y=\frac{{x}^{3}}{3}\) solves \({y}^{'}={x}^{2}\)

22. \(y=2{e}^{\text{-}x}+x-1\) solves \({y}^{'}=x-y\)

23. \(y={e}^{3x}-\frac{{e}^{x}}{2}\) solves \({y}^{'}=3y+{e}^{x}\)

24. \(y=\frac{1}{1-x}\) solves \({y}^{'}={y}^{2}\)

25. \(y={e}^{{x}^{2}\text{/}2}\) solves \({y}^{'}=xy\)

26. \(y=4+\text{ln}\ x\) solves \(x{y}^{'}=1\)

27. \(y=3-x+x\ \text{ln}\ x\) solves \({y}^{'}=\text{ln}\ x\)

28. \(y=2{e}^{x}-x-1\) solves \({y}^{'}=y+x\)

29. \(y={e}^{x}+\frac{\text{sin}\ x}{2}-\frac{\text{cos}\ x}{2}\) solves \({y}^{'}=\text{cos}\ x+y\)

30. \(y=\pi {e}^{\text{-}\text{cos}\ x}\) solves \({y}^{'}=y\ \text{sin}\ x\)

31. Find the particular solution to the differential equation \({y}^{'}=4{x}^{2}\) that passes through \((-3,-30),\) given that \(y=C+\frac{4{x}^{3}}{3}\) is a general solution.

32. Find the particular solution to the differential equation \({y}^{'}=3{x}^{3}\) that passes through \((1,4.75),\) given that \(y=C+\frac{3{x}^{4}}{4}\) is a general solution.

    Jiżvelaw it-tweġiba

    \(y=4+\frac{3{x}^{4}}{4}\)

33. Find the particular solution to the differential equation \({y}^{'}=3{x}^{2}y\) that passes through \((0,12),\) given that \(y=C{e}^{{x}^{3}}\) is a general solution.

34. Find the particular solution to the differential equation \({y}^{'}=2xy\) that passes through \((0,\frac{1}{2}),\) given that \(y=C{e}^{{x}^{2}}\) is a general solution.

    Jiżvelaw it-tweġiba

    \(y=\frac{1}{2}{e}^{{x}^{2}}\)

35. Find the particular solution to the differential equation \({y}^{'}={(2xy)}^{2}\) that passes through \((1,-\frac{1}{2}),\) given that \(y=-\frac{3}{C+4{x}^{3}}\) is a general solution.

36. Find the particular solution to the differential equation \({y}^{'}{x}^{2}=y\) that passes through \((1,\frac{2}{e}),\) given that \(y=C{e}^{\text{-}1\text{/}x}\) is a general solution.

    Jiżvelaw it-tweġiba

    \(y=2{e}^{\text{-}1\text{/}x}\)

37. Find the particular solution to the differential equation \(8\frac{dx}{dt}=-2\ \text{cos}(2t)-\text{cos}(4t)\) that passes through \((\pi ,\pi ),\) given that \(x=C-\frac{1}{8}\text{sin}(2t)-\frac{1}{32}\text{sin}(4t)\) is a general solution.

38. Find the particular solution to the differential equation \(\frac{du}{dt}=\text{tan}\ u\) that passes through \((1,\frac{\pi }{2}),\) given that \(u={\text{sin}}^{-1}({e}^{C+t})\) is a general solution.

    Jiżvelaw it-tweġiba

    \(u={\text{sin}}^{-1}({e}^{-1+t})\)

39. Find the particular solution to the differential equation \(\frac{dy}{dt}={e}^{(t+y)}\) that passes through \((1,0),\) given that \(y=\text{-}\text{ln}(C-{e}^{t})\) is a general solution.

40. Find the particular solution to the differential equation \({y}^{'}(1-{x}^{2})=1+y\) that passes through \((0,-2),\) given that \(y=C\frac{\sqrt{x+1}}{\sqrt{1-x}}-1\) is a general solution.

    Jiżvelaw it-tweġiba

    \(y=-\frac{\sqrt{x+1}}{\sqrt{1-x}}-1\)

Ipprova tiegħek stess

Parts of this page are adapted from OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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