2.3E: Linear First Order Equations (Exercises)
- Page ID
- 103469
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In Exercises 1-20 find the general solution.
1. \(y'+ay=0\) (\(a\)=constant)
2. \({dx\over dt}+x=e^{3t}\)
3. \(dy=(x^3-2xy)dx\)
4. \(\cos x{dy\over dx}+(\sin x)y=1\)
5. \(x^2y'+xy=1\)
6. \({dr\over d\theta}+r\sec \theta=\cos \theta\)
7. \( {y'+\left({1\over x}- 1\right)y=-{2\over x}}\)
8. \(y'+2xy=xe^{-x^2}\)
9. \({dy\over dx}+{2x\over 1+x^2}y={e^{-x}\over 1+x^2}\)
10. \( {y'+{1\over x}y={7\over x^2}+3}\)
11. \(y dx-4(x+y^6)dy=0\)
12. \( {dy\over dx}+{4\over x-1}y = {1\over (x-1)^5}+{\sin x\over (x-1)^4}\)
13. \(xy'+(1+2x^2)y=x^3e^{-x^2}\)
14. \( {xy'+2y={2\over x^2}+1}\)
15. \({dp\over dt}+(\tan t)p=\cos t\)
16. \( {(1+x)y'+2y={\sin x \over 1 + x}}\)
17. \((x-2)(x-1)y'-(4x-3)y=(x-2)^3\)
18. \(y^2dx=(e^y-xy^2-2xy)dy\)
19. \({dy\over dx}+(2\sin x\cos x) y=e^{-\sin^2x}\)
20. \(x^2y'+3xy=e^x\)
In Exercises 21-40 solve the initial value problem.
21. \( {y'+\left({1+x\over x}\right)y=0,\quad y(1)=1}\)
22. \( {x{dy\over dx}+\left(1+{1\over\ln x}\right)y=0,\quad y(e)=1}\)
23. \( {xdy+(1+ x\cot x)ydx=0,\quad y\left({\pi\over 2} \right)=2}\)
24. \( {y'-\left({2x\over 1+x^2}\right)y=0,\quad y(0)=2}\)
25. \( {dx\over dt}+\frac{k}{t}x=0,\quad x(1)=3\quad(k=\text{constant})\)
26. \( y'+(\tan kx)y=0,\quad y(0)=2\quad (k=\text{constant})\)
27. \(y'+7y=e^{3x},\quad y(0)=0\)
28. \( {(1+x^2)dy=({2\over 1+x^2}-4xy)dx,\quad y(0)=1}\)
29. \( {xy'+3y={2\over x(1+x^2)},\quad y(-1)=0}\)
30. \( {{dy\over dx}+ (\cot x)y=\cos x,\quad y\left({\pi\over 2}\right)=1}\)
31. \( {y'+{1\over x}y={2\over x^2}+1,\quad y(-1)=0}\)
32. \( {(x-1)y'+3y={1\over (x-1)^3} + {\sin x\over (x-1)^2},\quad y(0)=1}\)
33. \(ydx=(8y^2-2x)dy,\quad y(3)=1\)
34. \(xy'-2y=-x^2,\quad y(1)=1\)
35. \(y'+2xy=x,\quad y(0)=3\)
36. \( {(x-1)y'+3y={1+(x-1)\sec^2x\over (x-1)^3},\quad y(0)=-1}\)
37. \( {(x+2)y'+4y={1+2x^2\over x(x+2)^3},\quad y(-1)=2}\)
38. \((t^2-1){dx\over dx}-2tx=t(t^2-1),\quad x(0)=4\)
39. \((x^2-5)dy=-2x(x^2-5-y)dx,\quad y(2)=7\)
40. \((x+1){dy\over dx}+y=\ln x, \quad y(1)=10\)
41. Experiments indicate that glucose is absorbed by the body at a rate proportional to the amount of glucose present in the bloodstream. Let \(\lambda\) denote the (positive) constant of proportionality. Now suppose glucose is injected into a patient’s bloodstream at a constant rate of \(r\) units per unit of time. Let \(G=G(t)\) be the number of units in the patient’s bloodstream at time \(t>0\). Then \[G'=-\lambda G+r,\nonumber\] where the first term on the right is due to the absorption of the glucose by the patient’s body and the second term is due to the injection. Determine \(G\) for \(t>0\), given that \(G(0)=G_0\). Also, find \(\lim_{t\to\infty}G(t)\).
42. Assume that all functions in this exercise are defined on a common interval \((a,b)\).
Prove: If \(y_1\) and \(y_2\) are solutions of
\[y'+p(x)y=f_1(x)\nonumber\]
and
\[y'+p(x)y=f_2(x)\nonumber\]
respectively, and \(c_1\) and \(c_2\) are constants, then \(y=c_1y_1+c_2y_2\) is a solution of
\[y'+p(x)y=c_1f_1(x)+c_2f_2(x).\nonumber\]
(This is the principle of superposition.)