Find the solution to the differential equation

subject to the initial conditions

Ask by Hodges Maxwell. in the United States

Mar 21,2025

Question

Find the solution to the differential equation $$ 2 \frac{d u}{d t}=u^{2} $$ subject to the initial conditions $$ u(0)=5 $$ $$ u= $$

UpStudy AI · Accepted Answer

We start with the differential equation
$$
2 \frac{du}{dt}=u^{2}.
$$
This can be rewritten as
$$
\frac{du}{dt}=\frac{1}{2}u^2.
$$

**Step 1. Separate the Variables**

Separate the variables $$u$$ and $$t$$ by writing
$$
\frac{du}{u^2}=\frac{1}{2}\,dt.
$$

**Step 2. Integrate Both Sides**

Integrate both sides:
$$
\int \frac{du}{u^2}=\int \frac{1}{2}\,dt.
$$
The integral on the left is
$$
\int u^{-2}\,du = -u^{-1},
$$
and the integral on the right is
$$
\frac{1}{2}t.
$$
Thus, we have
$$
-\frac{1}{u}=\frac{t}{2}+C,
$$
where $$C$$ is the constant of integration.

**Step 3. Solve for the Constant Using the Initial Condition**

The initial condition is $$u(0)=5$$. Substitute $$t=0$$ and $$u=5$$ into the equation:
$$
-\frac{1}{5}=\frac{0}{2}+C,
$$
which gives
$$
C=-\frac{1}{5}.
$$

**Step 4. Write the Particular Solution**

Substitute $$C$$ back into the integrated equation:
$$
-\frac{1}{u}=\frac{t}{2}-\frac{1}{5}.
$$
Multiply both sides by $$-1$$:
$$
\frac{1}{u}=-\frac{t}{2}+\frac{1}{5}=\frac{1}{5}-\frac{t}{2}.
$$
Now solve for $$u$$ by taking the reciprocal:
$$
u=\frac{1}{\frac{1}{5}-\frac{t}{2}}.
$$

**Step 5. Simplify the Expression**

Find a common denominator for the terms in the denominator:
$$
\frac{1}{5}-\frac{t}{2}=\frac{2-5t}{10}.
$$
Thus,
$$
u=\frac{1}{\frac{2-5t}{10}}=\frac{10}{2-5t}.
$$

The solution to the differential equation is
$$
u(t)=\frac{10}{2-5t}.
$$
The solution to the differential equation is $$ u(t) = \frac{10}{2 - 5t} $$.

Find the solution to the differential equation

subject to the initial conditions

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Find the solution to the differential equation subject to the initial conditions