Question

The differential equation whose solution is \[y = ax + b{e^x}:\]
a. \[(x - 1){y_2} - x{y_1} + y = 0\]
b. \[(x - 1){y_2} - x{y_1} = y\]
c. \[{x^2}{y_2} - x{y_1} + y = 0\]
d. \[{x^2}{y_2} + x{y_1} - y = 0\]

Answer 1

Hint: We can solve this type of differential equations by differentiating only. We first differentiate the equation with respect to x to eliminate the x from the first term. And then we differentiate that equation again with respect to x, this will give the value of constant b. Putting the values of a and b in the first equation will give us our needed result.

Complete step-by-step answer:
\[y = ax + b{e^x}\]…….(i)
Differentiating with respect to x we get,
 \[ \Rightarrow \]\[{y_1} = a + b{e^x}\] where, \[{y_1} = \dfrac{{dy}}{{dx}}\] ……….(ii)
Again, Differentiating with respect to x we get,
 \[ \Rightarrow \] \[{y_2} = b{e^x}\] where, \[{y_2} = \dfrac{{{d^2}y}}{{d{x^2}}}\]
Then,
\[ \Rightarrow \] \[b = {y_2}{e^{ - x}}\]
Substituting the value of b in equation (ii), we get,
\[ \Rightarrow \] \[{y_1} = a + {y_2}{e^{ - x}}.{e^x}\]
On simplification we get,
 \[ \Rightarrow {y_1} = a + {y_2}\]
 \[ \Rightarrow a = {y_1} - {y_2}\]
Again, putting the value of a and b in equation (i), we get,
 \[y = ({y_1} - {y_2})x + {y_2}{e^{ - x}}.{e^x}\]
On simplification we get,
 \[ \Rightarrow y = x{y_1} - x{y_2} + {y_2}\]
On taking \[{y_2}\] common from last two terms we get,
 \[ \Rightarrow y = x{y_1} - {y_2}(x - 1)\]
On rearranging we get,
 \[ \Rightarrow (x - 1){y_2} - x{y_1} + y = 0\]
So, we have our answer as option a, \[(x - 1){y_2} - x{y_1} + y = 0\]

Note: In this type of problems the target of the problem is to find the values of distinct constants a and b. Then putting the values of a and b into our given equation will give us our desired result.
Basically we can say that we are eliminating the constants to get a differential equation.