Question

The D.E. whose solution is $y = ax + b{e^x}$ :

& A)\,\,\,(x - 1){y_2} - x{y_1} + y = 0 \cr & B)\,\,\,(x - 1){y_2} + x{y_1} = y \cr & C)\,\,\,{x^2}{y_2} - x{y_1} + y = 0 \cr & D)\,\,\,{x^2}{y_2} + x{y_1} - y = 0 \cr} $$

Answer 1

We have to take the derivative of y w.r.t. x twice so that we can get at first ${y_1} = \dfrac{{dy}}{{dx}},Ist\,order\,derivative\,of\,y\,\,w.r.t.\,x$ .Then ${y_2} = \dfrac{{{d^2}y}}{{d{x^2}}},2nd\,order\,derivative\,of\,y\,\,w.r.t.\,x$ .Observing to the options,we can form the required differential equation easily.

Complete step by step solution:
Step1: $Given,y = ax + b{e^x}$ .
Step2: Taking first order derivative
\eqalign{ & {y_1} = \dfrac{d}{{dx}}(ax + b{e^x}) \cr & \,\,\,\,\,\,\, = a + b{e^x} \cr}

Step3: Taking 2nd order derivative,
\eqalign{ & {y_2}\,\, = \dfrac{{{d^2}y}}{{d{x^2}}} \cr & \,\,\,\,\,\,\, = \dfrac{d}{{dx}}({y_1}) \cr & \,\,\,\,\,\,\, = \dfrac{d}{{dx}}(a + b{e^x}) \cr & \,\,\,\,\,\,\,\, = b{e^x} \cr}

Step4: Find the values of a and b. From step-2 and step-3,
$b = {e^{ - x}}{y_2}\,\,\,and\,\,\,a = {y_1} - {y_2}$
Step5: putting those values of a and b in $y = ax + b{e^x}$,
\eqalign{ & y = ({y_1} - {y_2})x + {e^{ - x}}{y_2}{e^x} \cr & or,y = ({y_1} - {y_2})x + {y_2} \cr & or,(x - 1){y_2} - x{y_1} + y = 0 \cr}

Hence option A) is correct here.

Note:
For quick process(applicable for MCQ only),after step 3, we have to form the differential equation.Putting the values of in L.H.S. of option A), we get
\eqalign{ & (x - 1){y_2} - x{y_1} + y \cr & = (x - 1)b{e^x} - x(a + b{e^x}) + ax + b{e^x} \cr & = b{e^x}(x - 1 - x + 1) - ax + ax \cr & = 0 \cr & Then,\,\,(x - 1){y_2} - x{y_1} + y = 0 \cr} .
If we see that option A) is false then we have to check if the second option is correct.In a similar way, we have to proceed further to check other options.