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NCERT Exemplar · Q12

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

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This is a first-order linear ODE solved using an integrating factor. The solution satisfying y(0)=−1y(0)=-1 gives y(1)=−12y(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)y(t) that satisfies both the equation and the starting value, then evaluate it at t=1t=1.

The equation is:

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

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

  1. Rewrite in standard linear form

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

dydt−t1+t y=11+t\frac{dy}{dt} - \frac{t}{1+t}\, y = \frac{1}{1+t}

So P(t)=−t1+tP(t) = -\frac{t}{1+t} and Q(t)=11+tQ(t) = \frac{1}{1+t}.

  1. Find the integrating factor

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

Compute ∫P(t) dt=∫−t1+t dt\int P(t)\, dt = \int -\frac{t}{1+t}\, dt.

Simplify the integrand: −t1+t=−(1−11+t)=−1+11+t-\frac{t}{1+t} = -\left(1 - \frac{1}{1+t}\right) = -1 + \frac{1}{1+t}.

So:

∫P(t) dt=∫(−1+11+t)dt=−t+log⁡∣1+t∣+C\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=0C=0. Then:

μ(t)=e−t+log⁡∣1+t∣=e−t⋅elog⁡∣1+t∣=(1+t)e−t\mu(t) = e^{-t + \log|1+t|} = e^{-t} \cdot e^{\log|1+t|} = (1+t) e^{-t}

Since t>−1t > -1 in our domain, 1+t>01+t > 0, so absolute values drop.

The integrating factor is μ(t)=(1+t)e−t\mu(t) = (1+t)e^{-t}.

  1. Multiply the ODE by μ(t)\mu(t)

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

(1+t)e−tdydt−(1+t)e−t⋅t1+t y=(1+t)e−t⋅11+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⋅t1+t=te−t(1+t)e^{-t} \cdot \frac{t}{1+t} = t e^{-t}. The right side becomes e−te^{-t}.

So we have:

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

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

ddt[(1+t)e−ty]=(1+t)e−tdydt+[e−t−(1+t)e−t]y=(1+t)e−tdydt−te−ty\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:

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

  1. Integrate both sides

Integrate from 00 to tt (or indefinitely and then use initial condition):

∫0tdds[(1+s)e−sy(s)]ds=∫0te−sds\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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