To find the particular solution to the differential equation [tex]\( y' x^2 = y \)[/tex] that passes through the point [tex]\(\left(1, \frac{2}{e}\right)\)[/tex], we start with the given general solution [tex]\( y = C e^{-1 / x} \)[/tex].
1. Substitute the given point into the general solution:
The point given is [tex]\( (1, \frac{2}{e}) \)[/tex]. Here, [tex]\( x = 1 \)[/tex] and [tex]\( y = \frac{2}{e} \)[/tex]. Substitute these values into the general solution to determine the constant [tex]\( C \)[/tex].
[tex]\[
y = C e^{-1 / x}
\][/tex]
Substituting [tex]\( x = 1 \)[/tex] and [tex]\( y = \frac{2}{e} \)[/tex]:
[tex]\[
\frac{2}{e} = C e^{-1 / 1}
\][/tex]
2. Simplify the expression:
[tex]\[
\frac{2}{e} = C e^{-1}
\][/tex]
Note that [tex]\( e^{-1} \)[/tex] is equivalent to [tex]\( \frac{1}{e} \)[/tex], so substitute that in:
[tex]\[
\frac{2}{e} = C \cdot \frac{1}{e}
\][/tex]
3. Solve for the constant [tex]\( C \)[/tex]:
[tex]\[
\frac{2}{e} = \frac{C}{e}
\][/tex]
Multiply both sides by [tex]\( e \)[/tex]:
[tex]\[
2 = C
\][/tex]
So, we find that [tex]\( C = 2 \)[/tex].
4. Write the particular solution:
Substitute [tex]\( C = 2 \)[/tex] back into the general solution [tex]\( y = C e^{-1 / x} \)[/tex]:
[tex]\[
y = 2 e^{-1 / x}
\][/tex]
Therefore, the particular solution to the differential equation [tex]\( y' x^2 = y \)[/tex] that passes through the point [tex]\(\left(1, \frac{2}{e}\right)\)[/tex] is:
[tex]\[
y = 2 e^{-1 / x}
\][/tex]
