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Find the solution of the following initial value problems. $$y^{\prime}(t)=\frac{3}{t}+6 ; y(1)=8$$

Short Answer

Expert verified
Question: Determine the particular solution of the initial value problem given the first-order differential equation \(y'(t) = \frac{3}{t} + 6\) with an initial condition \(y(1) = 8\). Answer: The particular solution is \(y(t) = 3\ln|t| + 6t + 2\).

Step by step solution

01

Identify the differential equation and initial condition

The given differential equation is $$ y^{\prime}(t) = \frac{3}{t} + 6 \quad and \quad y(1) = 8 $$ The goal is to find a function \(y(t)\) such that its derivative satisfies the differential equation, and the function passes through the point \((1,8)\).
02

Solve the differential equation

To solve the differential equation, we will integrate both sides with respect to \(t\): $$ \int y^{\prime}(t) dt = \int (\frac{3}{t} + 6) dt $$ Now, integrate the right side: $$ \int(\frac{3}{t} + 6)dt = 3\int \frac{1}{t}dt + 6\int dt = 3\ln|t| + 6t + C, $$ where \(C\) is the constant of integration. Now we have the general solution of the given differential equation: $$ y(t) = 3\ln|t| + 6t + C $$
03

Use the initial condition to find the constant of integration

We are given that \(y(1) = 8\). Plug in the initial condition into the general solution and solve for the constant \(C\): $$ 8 = 3\ln|1| + 6(1) + C $$ Since \(\ln|1| = 0\), the equation becomes: $$ 8 = 6 + C $$ Solve for \(C\). \( C = 8 - 6 = 2\).
04

Find the particular solution

Now that we have found the constant \(C\), we can find the particular solution of the given initial value problem: $$ y(t) = 3\ln|t| + 6t + 2 $$ This is the solution of the initial value problem.

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