IDNLearn.com provides a comprehensive solution for all your question and answer needs. Get thorough and trustworthy answers to your queries from our extensive network of knowledgeable professionals.
Sagot :
Let's address these differential equations one at a time, and find both the general solutions and particular solutions that pass through the origin (0, 0).
### Part (i)
Given the differential equation:
[tex]\[ \frac{d \theta}{d t} = \sin t + \cos t \][/tex]
First, we will find the general solution. We can integrate both sides with respect to [tex]\( t \)[/tex]:
[tex]\[ \int \frac{d \theta}{d t} \, dt = \int (\sin t + \cos t) \, dt \][/tex]
This simplifies to:
[tex]\[ \theta(t) = \int \sin t \, dt + \int \cos t \, dt \][/tex]
Integrate each term separately:
[tex]\[ \theta(t) = -\cos t + \sin t + C \][/tex]
Hence, the general solution is:
[tex]\[ \theta(t) = \sin t - \cos t + C \][/tex]
To find the particular solution that passes through the origin, we use the initial condition [tex]\(\theta(0) = 0\)[/tex]:
[tex]\[ \theta(0) = \sin 0 - \cos 0 + C = 0 - 1 + C = -1 + C = 0 \][/tex]
Thus, [tex]\( C = 1 \)[/tex].
Therefore, the particular solution is:
[tex]\[ \theta(t) = \sin t - \cos t + 1 \][/tex]
### Part (ii)
Given the differential equation:
[tex]\[ \frac{d y}{d x} = \frac{1}{x^2 - 1} \][/tex]
First, decompose the right-hand side into partial fractions. The expression [tex]\( \frac{1}{x^2 - 1} \)[/tex] can be factored as [tex]\( \frac{1}{(x - 1)(x + 1)} \)[/tex]:
[tex]\[ \frac{1}{x^2 - 1} = \frac{A}{x - 1} + \frac{B}{x + 1} \][/tex]
Solving for [tex]\( A \)[/tex] and [tex]\( B \)[/tex], we get:
[tex]\[ 1 = A(x + 1) + B(x - 1) \][/tex]
Substituting [tex]\( x = 1 \)[/tex]:
[tex]\[ 1 = A(1 + 1) + B(1 - 1) \][/tex]
[tex]\[ 1 = 2A \][/tex]
[tex]\[ A = \frac{1}{2} \][/tex]
Substituting [tex]\( x = -1 \)[/tex]:
[tex]\[ 1 = A(-1 + 1) + B(-1 - 1) \][/tex]
[tex]\[ 1 = -2B \][/tex]
[tex]\[ B = -\frac{1}{2} \][/tex]
Therefore,
[tex]\[ \frac{1}{x^2 - 1} = \frac{1/2}{x - 1} - \frac{1/2}{x + 1} \][/tex]
Thus, the given differential equation becomes:
[tex]\[ \frac{d y}{d x} = \frac{1/2}{x - 1} - \frac{1/2}{x + 1} \][/tex]
We integrate both sides with respect to [tex]\( x \)[/tex]:
[tex]\[ \int \frac{d y}{d x} \, dx = \int \left( \frac{1/2}{x - 1} - \frac{1/2}{x + 1} \right) \, dx \][/tex]
This simplifies to:
[tex]\[ y(x) = \frac{1}{2} \int \frac{1}{x - 1} \, dx - \frac{1}{2} \int \frac{1}{x + 1} \, dx + C \][/tex]
Integrate each term separately:
[tex]\[ y(x) = \frac{1}{2} \ln |x - 1| - \frac{1}{2} \ln |x + 1| + C \][/tex]
Combining the logarithmic terms, we get:
[tex]\[ y(x) = \frac{1}{2} \ln \left| \frac{x - 1}{x + 1} \right| + C \][/tex]
Hence, the general solution is:
[tex]\[ y(x) = \frac{1}{2} \ln \left| \frac{x - 1}{x + 1} \right| + C \][/tex]
To find the particular solution that passes through the origin, we use the initial condition [tex]\( y(0) = 0 \)[/tex]:
[tex]\[ 0 = \frac{1}{2} \ln \left| \frac{0 - 1}{0 + 1} \right| + C \][/tex]
[tex]\[ 0 = \frac{1}{2} \ln \left| -1 \right| + C \][/tex]
[tex]\[ 0 = \frac{1}{2} \ln 1 + C \][/tex]
[tex]\[ 0 = 0 + C \][/tex]
Thus, [tex]\( C = 0 \)[/tex].
Therefore, the particular solution is:
[tex]\[ y(x) = \frac{1}{2} \ln \left| \frac{x - 1}{x + 1} \right| \][/tex]
### Part (i)
Given the differential equation:
[tex]\[ \frac{d \theta}{d t} = \sin t + \cos t \][/tex]
First, we will find the general solution. We can integrate both sides with respect to [tex]\( t \)[/tex]:
[tex]\[ \int \frac{d \theta}{d t} \, dt = \int (\sin t + \cos t) \, dt \][/tex]
This simplifies to:
[tex]\[ \theta(t) = \int \sin t \, dt + \int \cos t \, dt \][/tex]
Integrate each term separately:
[tex]\[ \theta(t) = -\cos t + \sin t + C \][/tex]
Hence, the general solution is:
[tex]\[ \theta(t) = \sin t - \cos t + C \][/tex]
To find the particular solution that passes through the origin, we use the initial condition [tex]\(\theta(0) = 0\)[/tex]:
[tex]\[ \theta(0) = \sin 0 - \cos 0 + C = 0 - 1 + C = -1 + C = 0 \][/tex]
Thus, [tex]\( C = 1 \)[/tex].
Therefore, the particular solution is:
[tex]\[ \theta(t) = \sin t - \cos t + 1 \][/tex]
### Part (ii)
Given the differential equation:
[tex]\[ \frac{d y}{d x} = \frac{1}{x^2 - 1} \][/tex]
First, decompose the right-hand side into partial fractions. The expression [tex]\( \frac{1}{x^2 - 1} \)[/tex] can be factored as [tex]\( \frac{1}{(x - 1)(x + 1)} \)[/tex]:
[tex]\[ \frac{1}{x^2 - 1} = \frac{A}{x - 1} + \frac{B}{x + 1} \][/tex]
Solving for [tex]\( A \)[/tex] and [tex]\( B \)[/tex], we get:
[tex]\[ 1 = A(x + 1) + B(x - 1) \][/tex]
Substituting [tex]\( x = 1 \)[/tex]:
[tex]\[ 1 = A(1 + 1) + B(1 - 1) \][/tex]
[tex]\[ 1 = 2A \][/tex]
[tex]\[ A = \frac{1}{2} \][/tex]
Substituting [tex]\( x = -1 \)[/tex]:
[tex]\[ 1 = A(-1 + 1) + B(-1 - 1) \][/tex]
[tex]\[ 1 = -2B \][/tex]
[tex]\[ B = -\frac{1}{2} \][/tex]
Therefore,
[tex]\[ \frac{1}{x^2 - 1} = \frac{1/2}{x - 1} - \frac{1/2}{x + 1} \][/tex]
Thus, the given differential equation becomes:
[tex]\[ \frac{d y}{d x} = \frac{1/2}{x - 1} - \frac{1/2}{x + 1} \][/tex]
We integrate both sides with respect to [tex]\( x \)[/tex]:
[tex]\[ \int \frac{d y}{d x} \, dx = \int \left( \frac{1/2}{x - 1} - \frac{1/2}{x + 1} \right) \, dx \][/tex]
This simplifies to:
[tex]\[ y(x) = \frac{1}{2} \int \frac{1}{x - 1} \, dx - \frac{1}{2} \int \frac{1}{x + 1} \, dx + C \][/tex]
Integrate each term separately:
[tex]\[ y(x) = \frac{1}{2} \ln |x - 1| - \frac{1}{2} \ln |x + 1| + C \][/tex]
Combining the logarithmic terms, we get:
[tex]\[ y(x) = \frac{1}{2} \ln \left| \frac{x - 1}{x + 1} \right| + C \][/tex]
Hence, the general solution is:
[tex]\[ y(x) = \frac{1}{2} \ln \left| \frac{x - 1}{x + 1} \right| + C \][/tex]
To find the particular solution that passes through the origin, we use the initial condition [tex]\( y(0) = 0 \)[/tex]:
[tex]\[ 0 = \frac{1}{2} \ln \left| \frac{0 - 1}{0 + 1} \right| + C \][/tex]
[tex]\[ 0 = \frac{1}{2} \ln \left| -1 \right| + C \][/tex]
[tex]\[ 0 = \frac{1}{2} \ln 1 + C \][/tex]
[tex]\[ 0 = 0 + C \][/tex]
Thus, [tex]\( C = 0 \)[/tex].
Therefore, the particular solution is:
[tex]\[ y(x) = \frac{1}{2} \ln \left| \frac{x - 1}{x + 1} \right| \][/tex]
Your presence in our community is highly appreciated. Keep sharing your insights and solutions. Together, we can build a rich and valuable knowledge resource for everyone. Find clear answers at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.
