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The function powsolve solves a linear differential equation for which initial conditions do not have to be specified.

All the initial conditions must be at zero.

Derivatives are denoted by applying $\mathrm{D}$ to the function name. For example, the second derivative of $y$ at 0 is $\mathrm{D}\left(\mathrm{D}\left(y\right)\right)\left(0\right)$.

The solution returned is a formal power series that represents the infinite series solution.

In some cases, after assigning the name a to the output from the powsolve command, you can enter the command a(_k) to output a recurrence relation for the power series solution. See examples below.

The command with(powseries,powsolve) allows the use of the abbreviated form of this command.

$\mathrm{with}\left(\mathrm{powseries}\right)\:$

$a≔\mathrm{powsolve}\left(\mathrm{diff}\left(y\left(x\right)\,x\right)=y\left(x\right)\,y\left(0\right)=1\right)\:$

$\mathrm{tpsform}\left(a\,x\right)$

$1+x+\frac{1}{2}{x}^2+\frac{1}{6}{x}^3+\frac{1}{24}{x}^4+\frac{1}{120}{x}^5+O\left(x^6\right)$

$a\left(\mathrm{\_k}\right)$

$\frac{a\left(\mathrm{\_k}-1\right)}{\mathrm{\_k}}$

$v≔\mathrm{powsolve}\left(\left\{\mathrm{diff}\left(y\left(x\right)\,\mathrm{`\$`}\left(x\,4\right)\right)=y\left(x\right)\,y\left(0\right)=\frac{3}{2}\,\mathrm{D}\left(y\right)\left(0\right)=-\frac{1}{2}\,\mathrm{D}\left(\mathrm{D}\left(y\right)\right)\left(0\right)=-\frac{3}{2}\,\mathrm{D}\left(\mathrm{D}\left(\mathrm{D}\left(y\right)\right)\right)\left(0\right)=\frac{1}{2}\right\}\right)\:$

$\mathrm{tpsform}\left(v\,x\right)$

$\frac{3}{2}-\frac{1}{2}x-\frac{3}{4}{x}^2+\frac{1}{12}{x}^3+\frac{1}{16}{x}^4-\frac{1}{240}{x}^5+O\left(x^6\right)$