Question

Which of the following is true for y(x) that satisfies the differential equation \[\dfrac{{dy}}{{dx}} = xy - 1 + x - y;y(0) = 0\] ?
A. \[y{\rm{ }}\left( 1 \right){\rm{ }} = {\rm{ }}1\]
B. \[y{\rm{ }}\left( 1 \right){\rm{ }} = {\rm{ }}{e^{\dfrac{1}{2}}}\;-{\rm{ }}1\]
C.\[y{\rm{ }}\left( 1 \right){\rm{ }} = {\rm{ }}{e^{\dfrac{1}{2}}}\;-{\rm{ }}{e^{ - \dfrac{1}{2}}}\]
D. \[y{\rm{ }}\left( 1 \right){\rm{ }} = {\rm{ }}{e^{ - \dfrac{1}{2}}}\;-{\rm{ }}1\]

Answer 1

Hint: Use separation of variable method to separate the variables then integrate the function \[\dfrac{{dy}}{{y + 1}} = (x - 1)dx\] to obtain y from the given differential equation, then use the initial condition \[y(0) = 0\] to obtain the value of the constant C. Then substitute 1 for x in the equation \[\log \left| {y + 1} \right| = \dfrac{{{x^2}}}{2} - x\] to obtain \[y(1)\].

Formula used:
1. \[\int {\dfrac{{dx}}{{x - a}} = \log \left| {x - a} \right|} \]
2. \[\int {xdx = \dfrac{{{x^2}}}{2}} \]
3. \[\int {dx = x} \]

Complete step by step solution:
It is given that \[\dfrac{{dy}}{{dx}} = xy - 1 + x - y;y(0) = 0.\]
Now,
\[\dfrac{{dy}}{{dx}} = xy - 1 + x - y\]
\[\dfrac{{dy}}{{dx}} = xy + x - y - 1\]
\[\dfrac{{dy}}{{dx}} = x(y + 1) - 1(y + 1)\]
\[\dfrac{{dy}}{{dx}} = (x - 1)(y + 1)\]
\[\dfrac{{dy}}{{y + 1}} = (x - 1)dx\]
Integrate the equation \[\dfrac{{dy}}{{y + 1}} = (x - 1)dx\] to obtain the value of y.
\[\int {\dfrac{{dy}}{{y + 1}} = \int {(x - 1)dx} } \]
\[\log \left| {y + 1} \right| = \dfrac{{{x^2}}}{2} - x + C\], where C is integrating constant.
Substitute 0 for x and y in the equation \[\log \left| {y + 1} \right| = \dfrac{{{x^2}}}{2} - x + C\] to obtain the value of C.
So,
\[\begin{array}{l}\log \left| {0 + 1} \right| = \dfrac{{{0^2}}}{2} - 0 + C\\ \Rightarrow \log 1 = C\\ \Rightarrow C = 0\end{array}\]
Therefore, the equation is \[\log \left| {y + 1} \right| = \dfrac{{{x^2}}}{2} - x\]
Substitute 1 for x in the equation \[\log \left| {y + 1} \right| = \dfrac{{{x^2}}}{2} - x\] and obtain the value.
Hence,
\[\log \left| {y + 1} \right| = \dfrac{{{1^2}}}{2} - 1\]
\[ = - \dfrac{1}{2}\]
\[ \Rightarrow y + 1 = {e^{ - \dfrac{1}{2}}}\]
\[ \Rightarrow y = {e^{ - \dfrac{1}{2}}} - 1\]

The correct option is D.

Note: From the equation \[\log \left| {y + 1} \right| = \dfrac{{{x^2}}}{2} - x\] students can directly find the value of y as \[y = {e^{\dfrac{{{x^2}}}{2} - x}} - 1\], then substitute 1 for x in the equation \[y = {e^{\dfrac{{{x^2}}}{2} - x}} - 1\]and obtain the answer \[y = {e^{ - \dfrac{1}{2}}} - 1\].