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Step-by-Step Solution
Step 1: Write down the given differential equation
The problem provides the differential equation:
$
\frac{dy}{dx}
= (y + 1)\Big((y + 1)e^{\frac{x^2}{2}} - x\Big)
$
with the initial condition:
$
y(2) = 0.
$
Step 2: Introduce a suitable substitution
To simplify the expression, define
$
z = \frac{1}{y + 1}.
$
Then we will find an equation for
$
z.
$
Step 3: Rewrite the differential equation in terms of z
First, notice that
$
\frac{d}{dx}\Big(\frac{1}{y+1}\Big)
= -\frac{1}{(y+1)^2} \frac{dy}{dx}.
$
Hence,
$
-\frac{1}{(y+1)^2}\,\frac{dy}{dx} = \frac{dz}{dx}.
$
Given
$
\frac{dy}{dx}
= (y + 1)\Big((y + 1)e^{\frac{x^2}{2}} - x\Big),
$
multiplying both sides by
$
-\frac{1}{(y+1)^2}
$
gives
$
-\frac{1}{(y+1)^2}\,\frac{dy}{dx}
- x\left(\frac{1}{y+1}\right)
= -\,e^{\frac{x^2}{2}}.
$
In terms of
$
z = \frac{1}{y+1},
$
this becomes
$
\frac{dz}{dx} - x\,z = -\,e^{\frac{x^2}{2}}.
$
Step 4: Identify the linear differential equation in z
The equation
$
\frac{dz}{dx} - x\,z = -\,e^{\frac{x^2}{2}}
$
is a first-order linear differential equation of the form
$
\frac{dz}{dx} + P(x)\,z = Q(x),
$
where
$
P(x) = -x
$
and
$
Q(x) = -\,e^{\frac{x^2}{2}}.
$
Step 5: Compute the integrating factor (I.F.)
The integrating factor is given by
$
\text{I.F.}
= e^{\int P(x)\,dx}
= e^{\int -x\,dx}
= e^{-\frac{x^2}{2}}.
$
Step 6: Multiply through by the integrating factor and integrate
Multiplying the differential equation
$
\frac{dz}{dx} - x\,z = -\,e^{\frac{x^2}{2}}
$
by
$
e^{-\frac{x^2}{2}}
$
gives:
$
e^{-\frac{x^2}{2}} \frac{dz}{dx} - x\,z\,e^{-\frac{x^2}{2}}
= -\,e^{-\frac{x^2}{2}} e^{\frac{x^2}{2}}.
$
The left-hand side simplifies to
$
\frac{d}{dx}\Big(z\, e^{-\frac{x^2}{2}}\Big).
$
On the right-hand side,
$
-\,e^{-\frac{x^2}{2}} e^{\frac{x^2}{2}} = -1.
$
Thus,
$
\frac{d}{dx}\Big(z\, e^{-\frac{x^2}{2}}\Big) = -1.
$
Integrating both sides,
$
z\, e^{-\frac{x^2}{2}}
= -\int 1 \,dx
= -x + C,
$
where
$
C
$
is the constant of integration.
Step 7: Express z and hence y
Recall
$
z = \frac{1}{y + 1},
$
so
$
z\, e^{-\frac{x^2}{2}}
= -x + C
\quad \Longrightarrow \quad
\frac{1}{y+1} \, e^{-\frac{x^2}{2}}
= -x + C.
$
Therefore,
$
y + 1
= \frac{e^{-\frac{x^2}{2}}}{-x + C}.
$
Step 8: Use the initial condition to find C
The problem states
$
y(2) = 0.
$
So at
$
x = 2,
\, y = 0
$
and
$
y + 1 = 1.
$
Substituting
$
x = 2
$
and
$
y + 1 = 1
$
into
$
\frac{e^{-\frac{x^2}{2}}}{-x + C},
$
we get
$
1 = \frac{e^{-2}}{-2 + C}.
$
Hence,
$
-2 + C = e^{-2},
$
or
$
C = 2 + e^{-2}.
$
Step 9: Write the general solution for y
Substituting
$
C = 2 + e^{-2}
$
back into the expression for
$
y + 1,
$
we obtain:
$
y + 1
= \frac{e^{-\frac{x^2}{2}}}{-x + (2 + e^{-2})}
= \frac{e^{-\frac{x^2}{2}}}{(2 + e^{-2}) - x}.
$
Step 10: Evaluate y + 1 at x = 1
Set
$
x = 1.
$
Then
$
y + 1
= \frac{e^{-\frac{1^2}{2}}}{(2 + e^{-2}) - 1}
= \frac{e^{-\frac{1}{2}}}{1 + e^{-2}}
= \frac{e^{-\frac{1}{2}}}{\frac{e^2 + 1}{e^2}}
= \frac{e^{-\frac{1}{2}} \cdot e^2}{e^2 + 1}
= \frac{e^{\frac{3}{2}}}{e^2 + 1}.
$
Step 11: Compute $dy/dx$ at x = 1
We use the original differential equation:
$
\frac{dy}{dx}
= (y + 1)\Big((y + 1)e^{\frac{x^2}{2}} - x\Big).
$
At
$
x = 1,
$
we have
$
y + 1 = \frac{e^{\frac{3}{2}}}{e^2 + 1}.
$
So
$
(y + 1)e^{\frac{1^2}{2}}
= \frac{e^{\frac{3}{2}}}{e^2 + 1} \cdot e^{\frac{1}{2}}
= \frac{e^{2}}{e^2 + 1}.
$
Thus,
$
\Big((y + 1)e^{\frac{x^2}{2}} - x\Big)\Big|_{x=1}
= \frac{e^2}{e^2 + 1} - 1
= \frac{e^2 - (e^2 + 1)}{e^2 + 1}
= \frac{-1}{e^2 + 1}.
$
Hence,
$
\frac{dy}{dx}\Bigg|_{x=1}
= \left(\frac{e^{\frac{3}{2}}}{e^2 + 1}\right)
\left(\frac{-1}{e^2 + 1}\right)
= -\,\frac{e^{\frac{3}{2}}}{(e^2 + 1)^2}.
$
Final Answer
$
\boxed{- \,\frac{e^{\frac{3}{2}}}{\big(e^2 + 1\big)^2}}
$
