Question

What is the solution of the differential equation \[{e^{\dfrac{{dy}}{{dx}}}} = \left( {x + 1} \right)\], \[y\left( 0 \right) = 3\]?
A. \[y = x\log x - x + 2\]
B. \[y = \left( {x + 1} \right)\log \left| {x + 1} \right| - x + 3\]
C. \[y = \left( {x + 1} \right)\log \left| {x + 1} \right| + x + 3\]
D. \[y = x\log x + x + 3\]
E. \[y = - \left( {x + 1} \right)\log \left| {x + 1} \right| + x + 3\]

Answer 1

Hint: We will take the logarithm of both sides of the differential equation. Then we separate the variables of the differential equation and integrate both sides of the equation and solve the integration by using by parts formula. To calculate the integrating constant, we will put the initial condition.

Formula Used: Part parts formula:
\[\int {uvdx = u\int {vdx} - \int {\left[ {\dfrac{d}{{dx}}u\int {vdx} } \right]dx} } \]
u is the first function and v is the second function.

Complete step by step solution: Given differential equation is:
\[{e^{\dfrac{{dy}}{{dx}}}} = \left( {x + 1} \right)\]
Taking logarithms on both sides:
\[ \Rightarrow \log \left( {{e^{\dfrac{{dy}}{{dx}}}}} \right) = \log \left( {x + 1} \right)\]
Applying the formula \[\log {e^a} = a\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \log \left( {x + 1} \right)\]
Separate the variables of the above equation:
\[ \Rightarrow dy = \log \left( {x + 1} \right)dx\]
Taking integrating both sides:
\[ \Rightarrow \int {dy} = \int {\log \left( {x + 1} \right)dx} \]
Applying by parts formula on the right side:
\[ \Rightarrow y = \log \left( {x + 1} \right)\int {dx} - \int {\left[ {\dfrac{d}{{dx}}\left( {\log \left( {x + 1} \right)} \right)\int {dx} } \right]dx} \]
\[ \Rightarrow y = \log \left( {x + 1} \right) \cdot x - \int {\left[ {\dfrac{1}{{x + 1}} \cdot x} \right]dx} \]
\[ \Rightarrow y = \log \left( {x + 1} \right) \cdot x - \int {\dfrac{x}{{x + 1}}dx} \]
Add and subtract 1 with numerator of \[\int {\dfrac{x}{{x + 1}}dx} \]
\[ \Rightarrow y = \log \left( {x + 1} \right) \cdot x - \int {\dfrac{{x + 1 - 1}}{{x + 1}}dx} \]
\[ \Rightarrow y = \log \left( {x + 1} \right) \cdot x - \int {dx} + \int {\dfrac{1}{{x + 1}}dx} \]
\[ \Rightarrow y = \log \left( {x + 1} \right) \cdot x - x + \log \left| {x + 1} \right| + c\]
\[ \Rightarrow y = \left( {x + 1} \right)\log \left( {x + 1} \right) - x + c\] ….(i)
Now putting x = 0 and y = 3:
\[ \Rightarrow 3 = \left( {0 + 1} \right)\log \left( {0 + 1} \right) - 0 + c\]
\[ \Rightarrow 3 = c\]
Putting c = 3 in equation (i)
\[ \Rightarrow y = \left( {x + 1} \right)\log \left( {x + 1} \right) - x + 3\]

Option ‘B’ is correct

Note: Students often do mistake when they put initial condition. They put x =3 and y=0 which is incorrect. We have to put x = 0 and y = 3 in the solution of the differential equation to find the value of the integrating constant.