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Chapter 10: Problem 17

Solve the following differential equations: $$ y^{\prime}=(y-3)^{2} \ln t $$

Short Answer

Expert verified
y = 3 - \frac{1}{t \ln t - t + C}

Step by step solution

01

Identify the type of differential equation

The given differential equation is \( y' = (y - 3)^2 \ln t \). This is a first-order differential equation.
02

Separate the variables

Rearrange the equation to separate the variables y and t: \[ \frac{dy}{(y - 3)^2} = \ln t \ dt \]
03

Integrate both sides

Integrate both sides of the equation: \[ \int \frac{dy}{(y - 3)^2} = \int \ln t \ dt \] The left side can be integrated by recognizing it as a simple power rule integral: \[ \int (y - 3)^{-2} \ dy = \ \frac{-1}{y-3} + C_1 \] For the right side, use integration by parts where \ u = \ln t \ and \ dv = dt \: \[ \int \ln t \ dt = t \ln t - t + C_2 \] Combining both integrals, we get: \[ \frac{-1}{y - 3} = t \ln t - t + C \]
04

Solve for y

Rearrange the equation to solve for y: \[ \frac{-1}{y - 3} = t \ln t - t + C \] Taking the reciprocal of both sides: \[ y - 3 = \frac{-1}{t \ln t - t + C} \] Finally, solve for y: \[ y = 3 - \frac{1}{t \ln t - t + C} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Separation of Variables
Separation of variables is a powerful technique. We can often use it to solve first-order differential equations. This method involves rearranging the differential equation so that each variable and its differential are on opposite sides of the equation. In our example, the given differential equation is:
\( y' = (y - 3)^2 \, \ln t \)
We need to separate the variables \(y\) and \(t\) to integrate each side individually. Rearranging:
\[ \frac{dy}{(y - 3)^2} = \ln t \, dt \]
Now, both sides are ready for integration. Separating variables helps us simplify the equation step-by-step. This approach allows us to convert a complex relationship into manageable parts.
Integration by Parts
Integration by parts is a useful technique derived from the product rule of differentiation. It's helpful when integrating products of two functions. The basic formula for integration by parts is:
\[\int u \, dv = uv - \int v \, du\] In our equation, we need to integrate \( \ln t \) with respect to \( t \). We set:
\( u = \ln t \)
\( dv = dt \)
Then, differentiate \(u\) and integrate \(dv\):
\( du = \frac{1}{t} \, dt \)
\( v = t \)
Substitute into the integration by parts formula:
\[\int \ln t \, dt = t \ln t - \int t \, \frac{1}{t} \, dt \]
This simplifies to:
\[\int \ln t \, dt = t \ln t - t + C \]
This integration result is crucial for proceeding further.
Solving Differential Equations
Solving differential equations often involves combining techniques to find a general solution. Once we've separated variables and integrated each side, we need to solve for the dependent variable. In our equation, after integration, we have:
\[\frac{-1}{y - 3} = t \ln t - t + C\]
To solve for \(y\), take the reciprocal of both sides:
\[ y - 3 = \frac{-1}{t \ln t - t + C}\]
Finally, isolate \(y\):
\[ y = 3 - \frac{1}{t \ln t - t + C} \]
This yields the solution to the given first-order differential equation. Combining the methods of separating variables, integration, and isolation, we arrive at the final answer.

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Most popular questions from this chapter

When a red-hot steel rod is plunged in a bath of water that is kept at a constant temperature \(10^{\circ} \mathrm{C}\), the temperature of the rod at time \(t\), \(f(t)\), satisfies the differential equation $$ y^{\prime}=k[10-y] $$ where \(k>0\) is a constant of proportionality. Determine \(f(t)\) if the initial temperature of the rod is \(f(0)=350^{\circ} \mathrm{C}\) and \(k=.1\).

A person planning for her retirement arranges to make continuous deposits into a savings account at the rate of $$\$ 3600$$ per year. The savings account earns \(5 \%\) interest compounded continuously. (a) Set up a differential equation that is satisfied by \(f(t)\), the amount of money in the account at time \(t\). (b) Solve the differential equation in part (a), assuming that \(f(0)=0\), and determine how much money will be in the account at the end of 25 years.

Use Euler's method with \(n=4\) to approximate the solution \(f(t)\) to \(y^{\prime}=2 t-y+1, y(0)=5\) for \(0 \leq t \leq 2\) Estimate \(f(2)\).

Solve the given equation using an integrating factor. Take \(t>0\). $$ (1+t) y^{\prime}+y=-1 $$

Find an integrating factor for each equation. Take \(t>0\). $$ y^{\prime}+t y=6 t $$
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