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Question: Solve \(({x^2}{D^2} + 3xD + 1)y = \dfrac{1}{{{{(1 - x)}^2}}}\)...

Solve (x2D2+3xD+1)y=1(1−x)2({x^2}{D^2} + 3xD + 1)y = \dfrac{1}{{{{(1 - x)}^2}}}

Explanation

Solution

For solving this particular question, we have to let x=ez,z=ln⁡x,x>0,ddz≡θx = {e^z},z = \ln x,x > 0,\dfrac{d}{{dz}} \equiv \theta ,
After this our expression becomes less complex and we are able to proceed further , lastly we have to consider the concept of general solution.

Complete step by step solution:
It is given ,
(x2D2+3xD+1)y=1(1−x)2................(1)({x^2}{D^2} + 3xD + 1)y = \dfrac{1}{{{{(1 - x)}^2}}}................(1)
Now, let x=ez,z=ln⁡x,x>0,ddz≡θx = {e^z},z = \ln x,x > 0,\dfrac{d}{{dz}} \equiv \theta
⇒xD≡θ,x2D2≡θ(θ−1)\Rightarrow xD \equiv \theta ,{x^2}{D^2} \equiv \theta (\theta - 1)
Now, substitute all these in equation (1)(1) ,
We will get the following,
(θ(θ−1)+3θ+1)y=1(1−ez)2(\theta (\theta - 1) + 3\theta + 1)y = \dfrac{1}{{{{(1 - {e^z})}^2}}}
⇒(θ2+2θ+1)y=1(1−ez)2...............(2)\Rightarrow ({\theta ^2} + 2\theta + 1)y = \dfrac{1}{{{{(1 - {e^z})}^2}}}...............(2)
General solution is given by ,
y=yc+ypy = {y_c} + {y_p}
Now, we know that ,
yc=c1e−z+c2ze−z yp=1(θ2+2θ+1)(1−ez)2  {y_c} = {c_1}{e^{ - z}} + {c_2}z{e^{ - z}} \\\ {y_p} = \dfrac{1}{{({\theta ^2} + 2\theta + 1){{(1 - {e^z})}^2}}} \\\
=1θ+1(1θ+11(1−ez)2) =1θ+1(e−z∫1(1−ez)2ezdz) =1θ+1(e−z1−e−z) =e−z∫e−z1−e−zezdz  = \dfrac{1}{{\theta + 1}}\left( {\dfrac{1}{{\theta + 1}}\dfrac{1}{{{{(1 - {e^z})}^2}}}} \right) \\\ = \dfrac{1}{{\theta + 1}}\left( {{e^{ - z}}\int {\dfrac{1}{{{{(1 - {e^z})}^2}}}{e^z}dz} } \right) \\\ = \dfrac{1}{{\theta + 1}}\left( {\dfrac{{{e^{ - z}}}}{{1 - {e^{ - z}}}}} \right) \\\ = {e^{ - z}}\int {\dfrac{{{e^{ - z}}}}{{1 - {e^{ - z}}}}{e^z}dz} \\\
=−e−zln⁡∣e−z−1∣ yp=−e−zln⁡∣e−z−1∣ ∴y=c1e−z+c2ze−z−e−zln⁡∣e−z−1∣ ⇒y=c1x−1+c2(ln⁡x)x−1−x−1ln⁡∣x−1−1∣  = - {e^{ - z}}\ln \left| {{e^{ - z}} - 1} \right| \\\ {y_p} = - {e^{ - z}}\ln \left| {{e^{ - z}} - 1} \right| \\\ \therefore y = {c_1}{e^{ - z}} + {c_2}z{e^{ - z}} - {e^{ - z}}\ln \left| {{e^{ - z}} - 1} \right| \\\ \Rightarrow y = {c_1}{x^{ - 1}} + {c_2}(\ln x){x^{ - 1}} - {x^{ - 1}}\ln \left| {{x^{ - 1}} - 1} \right| \\\
Therefore, we get the required result.

Additional information:
A differential equation could be a mathematical equation for a function that relates the values of the function to its derivatives.
The highest power of the highest order derivative present within the differential equation is known as the degree of that equation.
The equation that is differential must be a polynomial equation for the degree to be defined.
A general solution to an ‘nth’ order differential equation could be a solution within which, the value of the constant within the solution, may vary.
A particular solution to an ‘nth’ order differential equation may be a solution within which, there's a selected value for the constant .
These sort of solutions occur within the initial value problems.
A function ‘y’ may be a solution of the given equation if the equation is satisfied by ‘y’ and its derivatives.

Note: Questions similar in nature as that of above can be approached in a similar manner and we can solve it easily. For solving this type of question, you must understand the concept that function ‘y’ may be a solution of the given equation if the equation is satisfied by ‘y’ and its derivatives.