Question

What is the solution of the differential equation \[y' - y = 1\] , if \[y\left( 0 \right) = - 1\]?
A. \[ - \exp \left( x \right)\]
B. \[ - \exp \left( { - x} \right)\]
C. \[ - 1\]
D. \[\exp \left( x \right) - 2\]

Answer 1

Hint: Here, a first-order differential equation is given. First, rewrite \[y'\] as \[\dfrac{{dy}}{{dx}}\] and simplify the differential equation. Then, integrate both sides with respect to the corresponding variables. After that, solve the integrals and find the solution of the differential equation. Then, substitute the values from the given equation \[y\left( 0 \right) = - 1\] in the solution of the differential equation and find the value of the integration constant \[c\]. In the end, substitute the value of the integration constant \[c\] in the solution of the differential equation to get the required answer.

Formula Used: \[\int {\dfrac{1}{x}} dx = \log x + c\]
\[\int n dx = nx + c\], where \[n\] is a number.
\[{e^{\log x}} = x\]

Complete step by step solution: The first-order differential equation is \[y' - y = 1\] and \[y\left( 0 \right) = - 1\].

Let’s solve the differential equation.
Rewrite \[y'\] as \[\dfrac{{dy}}{{dx}}\].
\[\dfrac{{dy}}{{dx}} - y = 1\]
Simplify the equation.
\[\dfrac{{dy}}{{dx}} = 1 + y\]
\[ \Rightarrow \dfrac{{dy}}{{1 + y}} = dx\]
Now integrate both sides with respect to the corresponding variables.
\[\int {\dfrac{{dy}}{{1 + y}}} = \int {dx} \]
Apply the integration formulas \[\int {\dfrac{1}{x}} dx = \log x + c\] and \[\int n dx = nx + c\].
We get,
\[\log \left( {1 + y} \right) = x + c\]
Take the above term as the power of \[e\].
\[{e^{\log \left( {1 + y} \right)}} = {e^{\left( {x + c} \right)}}\]
\[ \Rightarrow 1 + y = {e^x}{e^c}\] \[.....\left( 1 \right)\]

Now to calculate the value of the integration constant, substitute the values of the given equation \[y\left( 0 \right) = - 1\] in the equation \[\left( 1 \right)\].
We have, \[x = 0\] and \[y = - 1\]
Then,
\[1 + \left( { - 1} \right) = {e^0}{e^c}\]
\[ \Rightarrow 0 = 1 \times {e^c}\]
\[ \Rightarrow {e^c} = 0\]

Now substitute \[{e^c} = 0\] in the equation \[\left( 1 \right)\].
\[1 + y = {e^x}\left( 0 \right)\]
\[ \Rightarrow 1 + y = 0\]
\[ \Rightarrow y = - 1\]
Therefore, the solution of the differential equation \[y' - y = 1\] is \[y = - 1\].

Option ‘C’ is correct

Note: Sometimes students get confused and directly substitute the given values in the solution without resolving the logarithmic functions. If the numbers present in the logarithmic function are small, then we can directly calculate the value of the integration constant. But when numbers are large, then it became a complicated equation. So, to remove this situation take the logarithmic function as the power of the exponent.