Initial-value problems where the integral needs partial fractions or an arctangent. 60 problems with worked solutions; in 60 of them every
equation is proved by a computer algebra system.
Try each one before opening the solution. Every line is labeled with what checked it. How we verify
Solve \( \displaystyle \left(t^{2} + 2 t - 3\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(2) = 0 \) (for \( \displaystyle t > 1 \)).
Show the first step
Show the whole solution
Separate: dx = dt / (t**2 + 2*t - 3).
Reviewed
\[ \frac{1}{t^{2} + 2 t - 3} = - \frac{1}{4 t + 12} + \frac{1}{4 t - 4} \]
Partial fractions.✓ Proved
\[ \frac{d}{d t} \frac{\ln{\left(\frac{t - 1}{t + 3} \right)}}{4} = \frac{1}{t^{2} + 2 t - 3} \]
An antiderivative, checked by differentiating.✓ Proved
\[ - \left. \frac{\ln{\left(\frac{t - 1}{t + 3} \right)}}{4} \right|_{\substack{ t=2 }} + 0 = \frac{\ln{\left(5 \right)}}{4} \]
x(2) = 0 fixes the constant.✓ Proved
\[ \left. \frac{\ln{\left(\frac{5 \left(t - 1\right)}{t + 3} \right)}}{4} \right|_{\substack{ t=2 }} = 0 \]
The initial condition holds.✓ Proved
Answer \( x(t) = \frac{\ln{\left(\frac{5 \left(t - 1\right)}{t + 3} \right)}}{4} \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Solve \( \displaystyle \left(t^{2} - 9\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(5) = 0 \) (for \( \displaystyle t > 3 \)).
Show the first step
Show the whole solution
Separate: dx = dt / (t**2 - 9).
Reviewed
\[ \frac{1}{t^{2} - 9} = - \frac{1}{6 t + 18} + \frac{1}{6 t - 18} \]
Partial fractions.✓ Proved
\[ \frac{d}{d t} \frac{\ln{\left(\frac{t - 3}{t + 3} \right)}}{6} = \frac{1}{t^{2} - 9} \]
An antiderivative, checked by differentiating.✓ Proved
\[ - \left. \frac{\ln{\left(\frac{t - 3}{t + 3} \right)}}{6} \right|_{\substack{ t=5 }} + 0 = \frac{\ln{\left(2 \right)}}{3} \]
x(5) = 0 fixes the constant.✓ Proved
\[ \left. \frac{\ln{\left(\frac{4 \left(t - 3\right)}{t + 3} \right)}}{6} \right|_{\substack{ t=5 }} = 0 \]
The initial condition holds.✓ Proved
Answer \( x(t) = \frac{\ln{\left(\frac{4 \left(t - 3\right)}{t + 3} \right)}}{6} \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Solve \( \displaystyle \left(t^{2} - 6 t + 8\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(7) = 0 \) (for \( \displaystyle t > 4 \)).
Show the first step
Show the whole solution
Separate: dx = dt / (t**2 - 6*t + 8).
Reviewed
\[ \frac{1}{t^{2} - 6 t + 8} = - \frac{1}{2 t - 4} + \frac{1}{2 t - 8} \]
Partial fractions.✓ Proved
\[ \frac{d}{d t} \frac{\ln{\left(\frac{t - 4}{t - 2} \right)}}{2} = \frac{1}{t^{2} - 6 t + 8} \]
An antiderivative, checked by differentiating.✓ Proved
\[ - \left. \frac{\ln{\left(\frac{t - 4}{t - 2} \right)}}{2} \right|_{\substack{ t=7 }} + 0 = - \frac{\ln{\left(3 \right)}}{2} + \frac{\ln{\left(5 \right)}}{2} \]
x(7) = 0 fixes the constant.✓ Proved
\[ \left. \frac{\ln{\left(\frac{5 \left(t - 4\right)}{3 \left(t - 2\right)} \right)}}{2} \right|_{\substack{ t=7 }} = 0 \]
The initial condition holds.✓ Proved
Answer \( x(t) = \frac{\ln{\left(\frac{5 \left(t - 4\right)}{3 \left(t - 2\right)} \right)}}{2} \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Solve \( \displaystyle \left(t^{2} - 4 t\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(7) = 0 \) (for \( \displaystyle t > 4 \)).
Show the first step
Show the whole solution
Separate: dx = dt / (t**2 - 4*t).
Reviewed
\[ \frac{1}{t^{2} - 4 t} = \frac{1}{4 t - 16} - \frac{1}{4 t} \]
Partial fractions.✓ Proved
\[ \frac{d}{d t} \frac{\ln{\left(\frac{t - 4}{t} \right)}}{4} = \frac{1}{t^{2} - 4 t} \]
An antiderivative, checked by differentiating.✓ Proved
\[ - \left. \frac{\ln{\left(\frac{t - 4}{t} \right)}}{4} \right|_{\substack{ t=7 }} + 0 = - \frac{\ln{\left(3 \right)}}{4} + \frac{\ln{\left(7 \right)}}{4} \]
x(7) = 0 fixes the constant.✓ Proved
\[ \left. \frac{\ln{\left(\frac{7 \left(t - 4\right)}{3 t} \right)}}{4} \right|_{\substack{ t=7 }} = 0 \]
The initial condition holds.✓ Proved
Answer \( x(t) = \frac{\ln{\left(\frac{7 \left(t - 4\right)}{3 t} \right)}}{4} \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Solve \( \displaystyle \left(t^{2} - 3 t - 4\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(5) = 0 \) (for \( \displaystyle t > 4 \)).
Show the first step
Show the whole solution
Separate: dx = dt / (t**2 - 3*t - 4).
Reviewed
\[ \frac{1}{t^{2} - 3 t - 4} = - \frac{1}{5 t + 5} + \frac{1}{5 t - 20} \]
Partial fractions.✓ Proved
\[ \frac{d}{d t} \frac{\ln{\left(\frac{t - 4}{t + 1} \right)}}{5} = \frac{1}{t^{2} - 3 t - 4} \]
An antiderivative, checked by differentiating.✓ Proved
\[ - \left. \frac{\ln{\left(\frac{t - 4}{t + 1} \right)}}{5} \right|_{\substack{ t=5 }} + 0 = \frac{\ln{\left(6 \right)}}{5} \]
x(5) = 0 fixes the constant.✓ Proved
\[ \left. \frac{\ln{\left(\frac{6 \left(t - 4\right)}{t + 1} \right)}}{5} \right|_{\substack{ t=5 }} = 0 \]
The initial condition holds.✓ Proved
Answer \( x(t) = \frac{\ln{\left(\frac{6 \left(t - 4\right)}{t + 1} \right)}}{5} \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Solve \( \displaystyle \left(t^{2} + 4 t + 3\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(0) = 0 \) (for \( \displaystyle t > -1 \)).
Show the first step
Show the whole solution
Separate: dx = dt / (t**2 + 4*t + 3).
Reviewed
\[ \frac{1}{t^{2} + 4 t + 3} = - \frac{1}{2 t + 6} + \frac{1}{2 t + 2} \]
Partial fractions.✓ Proved
\[ \frac{d}{d t} \frac{\ln{\left(\frac{t + 1}{t + 3} \right)}}{2} = \frac{1}{t^{2} + 4 t + 3} \]
An antiderivative, checked by differentiating.✓ Proved
\[ - \left. \frac{\ln{\left(\frac{t + 1}{t + 3} \right)}}{2} \right|_{\substack{ t=0 }} + 0 = \frac{\ln{\left(3 \right)}}{2} \]
x(0) = 0 fixes the constant.✓ Proved
\[ \left. \frac{\ln{\left(\frac{3 \left(t + 1\right)}{t + 3} \right)}}{2} \right|_{\substack{ t=0 }} = 0 \]
The initial condition holds.✓ Proved
Answer \( x(t) = \frac{\ln{\left(\frac{3 \left(t + 1\right)}{t + 3} \right)}}{2} \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Solve \( \displaystyle \left(t^{2} - t - 6\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(6) = 0 \) (for \( \displaystyle t > 3 \)).
Show the first step
Show the whole solution
Separate: dx = dt / (t**2 - t - 6).
Reviewed
\[ \frac{1}{t^{2} - t - 6} = - \frac{1}{5 t + 10} + \frac{1}{5 t - 15} \]
Partial fractions.✓ Proved
\[ \frac{d}{d t} \frac{\ln{\left(\frac{t - 3}{t + 2} \right)}}{5} = \frac{1}{t^{2} - t - 6} \]
An antiderivative, checked by differentiating.✓ Proved
\[ - \left. \frac{\ln{\left(\frac{t - 3}{t + 2} \right)}}{5} \right|_{\substack{ t=6 }} + 0 = - \frac{\ln{\left(3 \right)}}{5} + \frac{3 \ln{\left(2 \right)}}{5} \]
x(6) = 0 fixes the constant.✓ Proved
\[ \left. \frac{\ln{\left(\frac{8 \left(t - 3\right)}{3 \left(t + 2\right)} \right)}}{5} \right|_{\substack{ t=6 }} = 0 \]
The initial condition holds.✓ Proved
Answer \( x(t) = \frac{\ln{\left(\frac{8 \left(t - 3\right)}{3 \left(t + 2\right)} \right)}}{5} \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Solve \( \displaystyle \left(t^{2} + 3 t\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(1) = 0 \) (for \( \displaystyle t > 0 \)).
Show the first step
Show the whole solution
Separate: dx = dt / (t**2 + 3*t).
Reviewed
\[ \frac{1}{t^{2} + 3 t} = - \frac{1}{3 t + 9} + \frac{1}{3 t} \]
Partial fractions.✓ Proved
\[ \frac{d}{d t} \frac{\ln{\left(\frac{t}{t + 3} \right)}}{3} = \frac{1}{t^{2} + 3 t} \]
An antiderivative, checked by differentiating.✓ Proved
\[ - \left. \frac{\ln{\left(\frac{t}{t + 3} \right)}}{3} \right|_{\substack{ t=1 }} + 0 = \frac{2 \ln{\left(2 \right)}}{3} \]
x(1) = 0 fixes the constant.✓ Proved
\[ \left. \frac{\ln{\left(\frac{4 t}{t + 3} \right)}}{3} \right|_{\substack{ t=1 }} = 0 \]
The initial condition holds.✓ Proved
Answer \( x(t) = \frac{\ln{\left(\frac{4 t}{t + 3} \right)}}{3} \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Solve \( \displaystyle \left(t^{2} + t - 2\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(4) = 0 \) (for \( \displaystyle t > 1 \)).
Show the first step
Show the whole solution
Separate: dx = dt / (t**2 + t - 2).
Reviewed
\[ \frac{1}{t^{2} + t - 2} = - \frac{1}{3 t + 6} + \frac{1}{3 t - 3} \]
Partial fractions.✓ Proved
\[ \frac{d}{d t} \frac{\ln{\left(\frac{t - 1}{t + 2} \right)}}{3} = \frac{1}{t^{2} + t - 2} \]
An antiderivative, checked by differentiating.✓ Proved
\[ - \left. \frac{\ln{\left(\frac{t - 1}{t + 2} \right)}}{3} \right|_{\substack{ t=4 }} + 0 = \frac{\ln{\left(2 \right)}}{3} \]
x(4) = 0 fixes the constant.✓ Proved
\[ \left. \frac{\ln{\left(\frac{2 \left(t - 1\right)}{t + 2} \right)}}{3} \right|_{\substack{ t=4 }} = 0 \]
The initial condition holds.✓ Proved
Answer \( x(t) = \frac{\ln{\left(\frac{2 \left(t - 1\right)}{t + 2} \right)}}{3} \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Solve \( \displaystyle \left(t^{2} + t\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(1) = 0 \) (for \( \displaystyle t > 0 \)).
Show the first step
Show the whole solution
Separate: dx = dt / (t**2 + t).
Reviewed
\[ \frac{1}{t^{2} + t} = - \frac{1}{t + 1} + \frac{1}{t} \]
Partial fractions.✓ Proved
\[ \frac{d}{d t} \ln{\left(\frac{t}{t + 1} \right)} = \frac{1}{t^{2} + t} \]
An antiderivative, checked by differentiating.✓ Proved
\[ - \left. \ln{\left(\frac{t}{t + 1} \right)} \right|_{\substack{ t=1 }} + 0 = \ln{\left(2 \right)} \]
x(1) = 0 fixes the constant.✓ Proved
\[ \left. \ln{\left(\frac{2 t}{t + 1} \right)} \right|_{\substack{ t=1 }} = 0 \]
The initial condition holds.✓ Proved
Answer \( x(t) = \ln{\left(\frac{2 t}{t + 1} \right)} \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Solve \( \displaystyle \left(t^{2} + 4\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(0) = 0 \).
Show the first step
Show the whole solution
Separate: dx = dt / (t**2 + 4).
Reviewed
1/(t² + 4) integrates to (1/2) arctan(t/2).
Reviewed
\[ \frac{d}{d t} \frac{\operatorname{atan}{\left(\frac{t}{2} \right)}}{2} = \frac{1}{t^{2} + 4} \]
An antiderivative, checked by differentiating.✓ Proved
\[ - \left. \frac{\operatorname{atan}{\left(\frac{t}{2} \right)}}{2} \right|_{\substack{ t=0 }} + 0 = 0 \]
x(0) = 0 fixes the constant.✓ Proved
\[ \left. \frac{\operatorname{atan}{\left(\frac{t}{2} \right)}}{2} \right|_{\substack{ t=0 }} = 0 \]
The initial condition holds.✓ Proved
Answer \( x(t) = \frac{\operatorname{atan}{\left(\frac{t}{2} \right)}}{2} \)
✓ Nihil obstat Lines: 3 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify