If $y(t)$ is a solution of $(1+t)\frac{dy}{dt}-ty=1$ and $y(0)=-1$, then show that $y(1)=-\frac{1}{2}$.

> [!TLDR] > This is a first-order linear ODE solved using an integrating factor. The solution satisfying $y(0)=-1$ gives $y(1) = -\frac12$. The problem gives us a differential equation and an initial condition — that’s an **Initial Value Problem (IVP)**. The goal is to find the specific function $y(t)$ that satisfies both the equation and the starting value, then evaluate it at $t=1$. The equation is: $$(1+t)\frac{dy}{dt} - t y = 1, \quad y(0) = -1$$ This is a **first-order linear ODE** in $y$. The standard form is $\frac{dy}{dt} + P(t) y = Q(t)$. Let’s rewrite it. --- 1. **Rewrite in standard linear form** Divide through by $(1+t)$ (valid for $t \neq -1$, and we only care about $t=0$ to $t=1$, so fine): $$\frac{dy}{dt} - \frac{t}{1+t}\, y = \frac{1}{1+t}$$ So $P(t) = -\frac{t}{1+t}$ and $Q(t) = \frac{1}{1+t}$. 2. **Find the integrating factor** The integrating factor is $\mu(t) = e^{\int P(t)\, dt}$. Compute $\int P(t)\, dt = \int -\frac{t}{1+t}\, dt$. Simplify the integrand: $-\frac{t}{1+t} = -\left(1 - \frac{1}{1+t}\right) = -1 + \frac{1}{1+t}$. So: $$\int P(t)\, dt = \int \left(-1 + \frac{1}{1+t}\right) dt = -t + \log|1+t| + C$$ We only need one antiderivative, so take $C=0$. Then: $$\mu(t) = e^{-t + \log|1+t|} = e^{-t} \cdot e^{\log|1+t|} = (1+t) e^{-t}$$ Since $t > -1$ in our domain, $1+t > 0$, so absolute values drop. > [!FORMULA] > The integrating factor is $\mu(t) = (1+t)e^{-t}$. 3. **Multiply the ODE by $\mu(t)$** Multiply the standard form by $\mu(t)$: $$(1+t)e^{-t} \frac{dy}{dt} - (1+t)e^{-t} \cdot \frac{t}{1+t}\, y = (1+t)e^{-t} \cdot \frac{1}{1+t}$$ Simplify the second term: $(1+t)e^{-t} \cdot \frac{t}{1+t} = t e^{-t}$. The right side becomes $e^{-t}$. So we have: $$(1+t)e^{-t} \frac{dy}{dt} - t e^{-t} y = e^{-t}$$ Notice the left side is exactly $\frac{d}{dt}\left[ (1+t)e^{-t} \, y \right]$. Check by differentiating: $$\frac{d}{dt}\left[ (1+t)e^{-t} y \right] = (1+t)e^{-t} \frac{dy}{dt} + \left[ e^{-t} - (1+t)e^{-t} \right] y = (1+

If y(t) is a solution of (1+t)(dy)/(dt)-ty=1 and y(0)=-1, then show that y(1)=-(1)/(2).

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This is a first-order linear ODE solved using an integrating factor. The solution satisfying $y(0)=-1$ gives $y(1) = -\frac12$ .

The problem gives us a differential equation and an initial condition — that’s an Initial Value Problem (IVP). The goal is to find the specific function $y(t)$ that satisfies both the equation and the starting value, then evaluate it at $t=1$ .

The equation is:

(1+t)\frac{dy}{dt} - t y = 1, \quad y(0) = -1

This is a first-order linear ODE in $y$ . The standard form is $\frac{dy}{dt} + P(t) y = Q(t)$ . Let’s rewrite it.

Rewrite in standard linear form

Divide through by $(1+t)$ (valid for $t \neq -1$ , and we only care about $t=0$ to $t=1$ , so fine):

\frac{dy}{dt} - \frac{t}{1+t}\, y = \frac{1}{1+t}

So $P(t) = -\frac{t}{1+t}$ and $Q(t) = \frac{1}{1+t}$ .

Find the integrating factor

The integrating factor is $\mu(t) = e^{\int P(t)\, dt}$ .

Compute $\int P(t)\, dt = \int -\frac{t}{1+t}\, dt$ .

Simplify the integrand: $-\frac{t}{1+t} = -\left(1 - \frac{1}{1+t}\right) = -1 + \frac{1}{1+t}$ .

So:

\int P(t)\, dt = \int \left(-1 + \frac{1}{1+t}\right) dt = -t + \log|1+t| + C

We only need one antiderivative, so take $C=0$ . Then:

\mu(t) = e^{-t + \log|1+t|} = e^{-t} \cdot e^{\log|1+t|} = (1+t) e^{-t}

Since $t > -1$ in our domain, $1+t > 0$ , so absolute values drop.

\mu(t) = (1+t)e^{-t}

Multiply the ODE by $\mu(t)$

Multiply the standard form by $\mu(t)$ :

(1+t)e^{-t} \frac{dy}{dt} - (1+t)e^{-t} \cdot \frac{t}{1+t}\, y = (1+t)e^{-t} \cdot \frac{1}{1+t}

Simplify the second term: $(1+t)e^{-t} \cdot \frac{t}{1+t} = t e^{-t}$ . The right side becomes $e^{-t}$ .

So we have:

(1+t)e^{-t} \frac{dy}{dt} - t e^{-t} y = e^{-t}

Notice the left side is exactly $\frac{d}{dt}\left[ (1+t)e^{-t} \, y \right]$ . Check by differentiating:

\frac{d}{dt}\left[ (1+t)e^{-t} y \right] = (1+t)e^{-t} \frac{dy}{dt} + \left[ e^{-t} - (1+t)e^{-t} \right] y = (1+t)e^{-t} \frac{dy}{dt} - t e^{-t} y

Yes, matches.

So the equation becomes:

\frac{d}{dt}\left[ (1+t)e^{-t} y \right] = e^{-t}

Integrate both sides

Integrate from $0$ to $t$ (or indefinitely and then use initial condition):

$\int_0^t \frac{d}{ds}\left[ (1+s)e^{-s} y(s) \right] ds = \int_0^t e^{-s} ds$ …

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Q.If $y(t)$ is a solution of $(1+t)\frac{dy}{dt}-ty=1$ and $y(0)=-1$ , then show that $y(1)=-\frac{1}{2}$ .

Concept understanding — Initial Value Problem

Initial Value Problem

The intuition: a rule of change plus a starting point

The precise statement

A worked example

Method: Linear IVP via integrating factor, then verifying a value

Steps

Common Mistakes

Q.If y(t)y(t)y(t) is a solution of (1+t)dydt−ty=1(1+t)\frac{dy}{dt}-ty=1(1+t)dtdy​−ty=1 and y(0)=−1y(0)=-1y(0)=−1, then show that y(1)=−12y(1)=-\frac{1}{2}y(1)=−21​.

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Q.If $y(t)$ is a solution of $(1+t)\frac{dy}{dt}-ty=1$ and $y(0)=-1$ , then show that $y(1)=-\frac{1}{2}$ .

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