Animated Solution for Mathematics - Differential Equations: If $y(t)$ is a solution of $(1+t)\frac{dy}{dt}-ty=1$ and $y(0)=-1$, then $y(1)$ is equal to

Welcome, future engineer! Today, we are going to dive into a problem that might look like a tangled mess of variables at first glance, but beneath the surface, it is a beautifully structured linear differential equation.

The problem asks us to solve $(1+t)\frac{dy}{dt}-ty=1$ with the initial condition $y(0)=-1$. Our mission is to find $y(1)$. Let's embark on this journey together.

In the world of differential equations, the first step is always to identify the 'type' of the equation. We see a $\frac{dy}{dt}$ term, a $y$ term, and some functions of $t$. This is the signature of a linear differential equation.

To solve it, we must bring it into the standard form: $\frac{dy}{dt}+P(t)y=Q(t)$. Currently, our equation is $(1+t)\frac{dy}{dt}-ty=1$.

The coefficient of $\frac{dy}{dt}$ is $(1+t)$, not $1$. So, we must divide the entire equation by $(1+t)$. This gives us:

Now, we can clearly see that $P(t)=-\frac{t}{1+t}$ and $Q(t)=\frac{1}{1+t}$.

Now, we need the magic ingredient: the Integrating Factor (I.F.). The formula is $I.F.=e^{\int P(t)dt}$.

Before we jump into the exponent, let's simplify $P(t)=-\frac{t}{1+t}$. Integrating this directly is a bit messy. Here is a classic JEE trick: add and subtract $1$ in the numerator.

This turns $P(t)$ into $\frac{1-(1+t)}{1+t}$, which simplifies to $\frac{1}{1+t}-1$. Now, the integration becomes a breeze:

Our Integrating Factor is $e^{\ln (1+t)-t}$. Using the laws of exponents, this becomes $e^{\ln (1+t)}\cdot e^{-t}$, which simplifies beautifully to $(1+t)e^{-t}$.

With our I.F. in hand, the general solution is given by $y\cdot (I.F.)=\int Q(t)\cdot (I.F.)dt+C$. Substituting our values, we get:

Notice how the $(1+t)$ terms cancel out perfectly on the right side! We are left with $\int e^{-t}dt+C$. The integral of $e^{-t}$ is simply $-e^{-t}$.

So, our general solution is $y(1+t)e^{-t}=-e^{-t}+C$.

We are almost there! We have a constant $C$ that we need to determine. We use the initial condition $y(0)=-1$.

Substituting $t=0$ and $y=-1$ into our equation: $(-1)(1+0)e^0=-e^0+C$. Since $e^0=1$, this simplifies to $-1=-1+C$, which means $C=0$.

How elegant! The constant vanishes, leaving us with the specific solution $y(1+t)e^{-t}=-e^{-t}$. Dividing both sides by $(1+t)e^{-t}$, we get:

Finally, we need to find $y(1)$. Substituting $t=1$ into our specific solution, we get:

We have navigated the complexity and arrived at a clean, simple answer. Remember, in JEE, the most complex-looking equations often have the most elegant solutions if you just follow the steps. Keep practicing, and you will master these patterns!