Can we solve 1st order q-differential equations using the usual methods of 1st order differential equations? For example, can we use integration factor method to solve this q-differential equations?_qy(x)=a(x)y(qx)+b(xqy(x)=a(x)y(qx)+b(x)

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hinonacfp · Accepted Answer

Step 1Here is how to get a functional equation since you asked. Starting with a direct reference from q-Derivative from Wolfram MathWorld:(d&amp;nbsp;dx&amp;nbsp;)qf(x)=f(x)−f(qx)x−qxTherefore:_q y(x)=a(x)y(qx)+b(xqy(x)=a(x)y(qx)+b(x)=y(qx)−y(x)x(q−1)=a(x)y(qx)+b(x)which works for&amp;nbsp;q≠1&amp;nbsp;since:limq→1_q y(xqy(x)=limq→1y(qx)−y(x)x(q−1)=limq→0y(x+q)−y(x)q=y′(x)=&amp;nbsp;&amp;nbsp;y(xy(x)=&amp;nbsp;dy&amp;nbsp;(x)&amp;nbsp;dx&amp;nbsp;but this limit is not needed when the q-Derivative is taken. Therefore we have our functional equation, rewritten:Dqy(x)=a(x)y(qx)+b(x)⇔y(qx)-y(x)x(q-1)=a(x)y(qx)+b(x)Here is another form which also gives a recursive solution:_q y(x)=a(x)y(qx)+b(x)iff y(qx)-y(xqy(x)=a(x)y(qx)+b(x)⇔y(qx)−y(x)=a(x)y(qx)x(q−1)+b(x)x(q−1)⇒y(x)=y(qx)(a(x)x(1−q)+1)−b(x)x(q−1)Please correct me and give me feedback!

Can we solve 1st order q-differential equations using

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