#### Example 4.1:

*Consider the following FDE*
$4x{D}^{1/2}{D}^{1/2}z-2(x+2\sqrt{x}){D}^{1/2}z+(\sqrt{x}+2)z=4x\sqrt{x}{e}^{\sqrt{x}},$

*we have $P=-(1+2/\sqrt{x}),\hspace{0.17em}\hspace{0.17em}Q=\frac{1}{\sqrt{x}}+\frac{2}{x},\hspace{0.17em}\hspace{0.17em}R=\sqrt{x}{e}^{\sqrt{x}}$ And $P+Q{x}^{1/2}=0$ obviously, so complete solution will be given by **z* = *uv*, where $u=\sqrt{x}$ is known function.

*Now to determine **v* we have the following differential equation
${D}^{1/2}{D}^{1/2}v+\left(\frac{1}{2}P+\frac{2}{u}{D}^{1/2}u\right){D}^{1/2}v=R/u,$
$\u27f9{D}^{1/2}{D}^{1/2}v+\left(-\frac{1}{2}-\frac{1}{\sqrt{x}}+\frac{2}{\sqrt{x}}{D}^{1/2}\sqrt{x}\right){D}^{1/2}v={e}^{\sqrt{x}},$
$\u27f9{D}^{1/2}{D}^{1/2}v+\left(-\frac{1}{2}-\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x}}\right){D}^{1/2}v={e}^{\sqrt{x}},$

*taking $q={D}^{1/2}v,$ we get*
$\frac{{d}^{1/2}q}{d{x}^{1/2}}-\frac{1}{2}q={e}^{\sqrt{x}}\u27f9I.F={e}^{\underset{1/2}{\int}-\frac{1}{2}d{x}^{1/2}}={e}^{\int -\frac{1}{2\sqrt{x}}dx}={e}^{-\sqrt{x}},$

*So*
$\begin{array}{rl}q{e}^{-\sqrt{x}}& =\underset{1/2}{\int}{e}^{-\sqrt{x}}{e}^{\sqrt{x}}d{x}^{1/2}+a\u27f9q=\frac{{d}^{1/2}v}{d{x}^{1/2}}\\ & ={e}^{\sqrt{x}}\int \frac{1}{\sqrt{x}}dx+a{e}^{\sqrt{x}},\end{array}$

*And therefore*,
$v=\int {e}^{\sqrt{x}}\left(2\sqrt{x}+a\right)\frac{1}{\sqrt{x}}dx+b=\int 2{e}^{\sqrt{x}}dx+2a\sqrt{x}+b$

*Hence*,
$z=(\int 2{e}^{\sqrt{x}}dx+2a\sqrt{x}+b){e}^{\sqrt{x}}$

*where **a*,*b* are constants.

#### Example 4.2:

*Consider the following FDE*
${D}^{1/3}{y}^{1/3}-\frac{1}{3}cot{x}^{1/3}{y}^{1/3}-\frac{1}{9}(1-cot{x}^{1/3})y={e}^{{x}^{1/3}}sin{x}^{1/3}.$

*Here we have*
$P=-cot{x}^{1/3},Q=cot{x}^{1/3}-1,R={e}^{{x}^{1/3}}sin{x}^{1/3}$
$1+P+Q=0\u27f9{e}^{{x}^{1/3}}$

The one part of C.F is ${e}^{{x}^{1/3}},$ Then now to determine *v* we have
$\begin{array}{r}u={e}^{{x}^{1/3}}\u27f9\frac{{d}^{1/3}}{d{x}^{1/3}}\frac{{d}^{1/3}v}{d{x}^{1/3}}+\left(\frac{1}{3}P+\frac{2}{u}\frac{{d}^{1/3}u}{d{x}^{1/3}}\right)\frac{{d}^{1/3}v}{d{x}^{1/3}}=sin{x}^{1/3},\end{array}$
$\begin{array}{r}\u27f9\frac{{d}^{1/3}}{d{x}^{1/3}}\frac{{d}^{1/3}v}{d{x}^{1/3}}+\left(-\frac{1}{3}cot{x}^{1/3}+\frac{2}{{e}^{{x}^{1/3}}}\frac{{d}^{1/3}u}{d{x}^{1/3}}\right)\frac{{d}^{1/3}v}{d{x}^{1/3}}=sin{x}^{1/3},\end{array}$
$\begin{array}{r}\u27f9\frac{{d}^{1/3}}{d{x}^{1/3}}\frac{{d}^{1/3}v}{d{x}^{1/3}}+\left(-\frac{1}{3}cot{x}^{1/3}+\frac{2}{{e}^{{x}^{1/3}}}{x}^{2/3}\frac{{d}^{}u}{d{x}^{}}\right)\frac{{d}^{1/3}v}{d{x}^{1/3}}=sin{x}^{1/3},\end{array}$
$\begin{array}{r}\u27f9\frac{{d}^{1/3}q}{d{x}^{1/3}}+\left(-\frac{1}{3}cot{x}^{1/3}+\frac{2}{3}\right)q=sin{x}^{1/3},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}q=\frac{{d}^{1/3}v}{d{x}^{1/3}},\end{array}$
$\begin{array}{r}I.F.={e}^{{\int}_{1/3}\frac{(2-cot{x}^{1/3})}{3}d{x}^{1/3}}={e}^{\int \frac{(2-cot{x}^{1/3})}{3{x}^{2/3}}dx}=\frac{{e}^{2{x}^{1/3}}}{sin{x}^{1/3}},\end{array}$
$\begin{array}{rl}\u27f9q\frac{{e}^{2{x}^{1/3}}}{sin{x}^{1/3}}& =3\frac{{e}^{2{x}^{1/3}}}{2}+c\u27f9q=\frac{{d}^{1/3}v}{d{x}^{1/3}}=3\frac{sin{x}^{1/3}}{2}\\ & +c{e}^{-2{x}^{1/3}}sin{x}^{1/3}\end{array}$
$\begin{array}{r}\u27f9v=\int \left(3\frac{sin{x}^{1/3}}{2}+c{e}^{-2{x}^{1/3}}sin{x}^{1/3}\right){x}^{-2/3}dx+{c}^{\prime}\end{array}$
$\begin{array}{r}=\frac{-9cos{x}^{1/3}}{2}-\frac{3}{5}c{e}^{-2{x}^{1/3}}(2sin{x}^{1/3}+cos{x}^{1/3})+{c}^{\prime}\end{array}$

*Hence*,
$\begin{array}{r}y=\left(\frac{-9cos{x}^{1/3}}{2}-\frac{3}{5}c{e}^{-2{x}^{1/3}}(2sin{x}^{1/3}+cos{x}^{1/3})+{c}^{\prime}\right){e}^{{x}^{1/3}}\end{array}$

#### Example 4.3:

*Consider the following fractional differential equation*
${D}^{\alpha}{D}^{\alpha}y-3\alpha {D}^{\alpha}y+2{\alpha}^{2}y=\frac{{e}^{{x}^{\alpha}}}{1+{e}^{{x}^{\alpha}}},\hspace{0.17em}\hspace{0.17em}0<\alpha \u2a7d1$

*Let us consider $y={e}^{m{x}^{\alpha}}$ as a solution, then auxiliary equation is*
${m}^{2}-3m+2=0\u27f9m=1,2.$

So
$C.F={c}_{1}{e}^{{x}^{\alpha}}+{c}_{2}{e}^{2{x}^{\alpha}}.$

*Take*
$u={e}^{{x}^{\alpha}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}v={e}^{2{x}^{\alpha}}$

*Then Wronskian*,
${W}_{\alpha}=\left|\begin{array}{rr}u& v\\ {u}^{\alpha}& {v}^{\alpha}\end{array}\right|=\left|\begin{array}{rr}{e}^{{x}^{\alpha}}& {e}^{2{x}^{\alpha}}\\ \alpha {e}^{{x}^{\alpha}}& 2\alpha {e}^{2{x}^{\alpha}}\end{array}\right|=\alpha {e}^{3{x}^{\alpha}}$
$P.I=Xu+Yv$

*where*
$\begin{array}{rl}X& ={\int}_{\alpha}\frac{-vR}{{W}_{\alpha}}d{x}^{\alpha}={\int}_{\alpha}\frac{-{e}^{2{x}^{\alpha}}{e}^{{x}^{\alpha}}}{\alpha {e}^{3{x}^{\alpha}}(1+{e}^{{x}^{\alpha}})}d{x}^{\alpha}\\ & =\int \frac{-{x}^{\alpha -1}}{\alpha (1+{e}^{{x}^{\alpha}})}dx;\end{array}$
$\begin{array}{rl}{x}^{\alpha}& =t\u27f9{x}^{\alpha -1}dx=\frac{dx}{\alpha}\u27f9\int \frac{-{e}^{-t}}{{\alpha}^{2}(1+{e}^{-t})}dt\\ & =\frac{1}{{\alpha}^{2}}log(1+{e}^{-{x}^{\alpha}})\end{array}$

*and*
$Y={\int}_{\alpha}\frac{uR}{{W}_{\alpha}}d{x}^{\alpha}={\int}_{\alpha}\frac{{e}^{{x}^{\alpha}}{e}^{{x}^{\alpha}}}{\alpha {e}^{3{x}^{\alpha}}(1+{e}^{{x}^{\alpha}})}d{x}^{\alpha}=\int \frac{{x}^{\alpha -1}}{\alpha {e}^{{x}^{\alpha}}(1+{e}^{{x}^{\alpha}})}dx$
$=\int \frac{1}{{\alpha}^{2}{e}^{t}(1+{e}^{t})}dt=\frac{1}{{\alpha}^{2}}log\left(\frac{{e}^{{x}^{\alpha}}}{1+{e}^{{x}^{\alpha}}}\right)$

*Hence, the complete solution*
$\begin{array}{rl}y=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}& {c}_{1}{e}^{{x}^{\alpha}}+{c}_{2}{e}^{2{x}^{\alpha}}+\left(\frac{1}{{\alpha}^{2}}log(1+{e}^{-{x}^{\alpha}})\right){e}^{{x}^{\alpha}}\\ & +\left(\frac{1}{{\alpha}^{2}}log\left(\frac{{e}^{{x}^{\alpha}}}{1+{e}^{{x}^{\alpha}}}\right)\right){e}^{2{x}^{\alpha}}\end{array}$

#### Example 4.4:

*Solve the following FDE*
$(1-{x}^{\beta}){D}^{\beta}{y}^{\beta}+\beta {x}^{\beta}{y}^{\beta}-{\beta}^{2}y=2({x}^{\beta}-1{)}^{2}{e}^{-{x}^{\beta}};0<\beta \u2a7d1.$

We have
${D}^{\beta}{y}^{\beta}+\beta \frac{{x}^{\beta}}{1-{x}^{\beta}}{y}^{\beta}-{\beta}^{2}\frac{1}{1-{x}^{\beta}}y=2\frac{(1-{x}^{\beta})}{{e}^{{x}^{\beta}}}$

so that
$P=\frac{{x}^{\beta}}{1-{x}^{\beta}},\hspace{0.17em}\hspace{0.17em}Q=\frac{1}{1-{x}^{\beta}},\hspace{0.17em}\hspace{0.17em}R=2\frac{1-{x}^{\beta}}{{e}^{{x}^{\beta}}}.$

*Now*,
$1+P+Q=0\u27f9{e}^{{x}^{\beta}}$

*is the one part of the complementary function and $P+Q{x}^{\beta}=0\u27f9{x}^{\beta}$ will be second independent solution and so complementary function is ${c}_{1}{e}^{{x}^{\beta}}+{c}_{2}{x}^{\beta}$*

*Set $u={e}^{{x}^{\beta}},v={x}^{\beta},$ so that Wronskian*
${W}_{\beta}=\left|\begin{array}{rr}u& v\\ {u}^{\beta}& {v}^{\beta}\end{array}\right|=\left|\begin{array}{rr}{e}^{{x}^{\beta}}& {x}^{\beta}\\ \beta {e}^{{x}^{\beta}}& \beta \end{array}\right|=\beta (1-{x}^{\beta}){e}^{{x}^{\beta}}$

*and for particular integral*,
${y}_{p}=Xu+Yv$
$\begin{array}{rl}X& ={\int}_{\beta}\frac{-vR}{{W}_{\beta}}d{x}^{\beta}=\int \frac{-2(1-{x}^{\beta}){x}^{\beta -1}{x}^{\beta}}{\beta {e}^{2{x}^{\beta}}(1-{x}^{\beta})}dx\\ & =-\int 2{x}^{\beta}{x}^{\beta -1}{e}^{-2{x}^{\beta}}dx\end{array}$
${x}^{\beta}=t\u27f9{x}^{\beta -1}dx=\frac{dt}{\beta}=-\int 2t{e}^{-2t}dt=\frac{1}{\beta {e}^{2{x}^{\beta}}}\left(\frac{1}{2}+{x}^{\beta}\right)$

*and*
$\begin{array}{rl}Y& ={\int}_{\beta}\frac{uR}{{W}_{\beta}}d{x}^{\beta}=\int \frac{2(1-{x}^{\beta}){x}^{\beta -1}{e}^{{x}^{\beta}}}{\beta {e}^{2{x}^{\beta}}(1-{x}^{\beta})}dx=\int 2{x}^{\beta -1}{e}^{-{x}^{\beta}}dx\\ & =\frac{-2{e}^{-{x}^{\beta}}}{\beta}\end{array}$

*Therefore, the complete solution*
$y=({c}_{1}+X){e}^{{x}^{\beta}}+({c}_{2}+Y){x}^{\beta}$

#### Example 4.5:

*Consider the following FDE*
${D}^{1/2}{D}^{1/2}y+y=tan2\sqrt{x}.$

*Let us assume that $y={e}^{m{x}^{1/2}}$as a solution of homogeneous part, then the corresponding*

A.E. is
${m}^{2}+4=0\Rightarrow m=\pm 2i$
$C.F=asin2\sqrt{x}+bcos2\sqrt{x},$

*choose*
$u=sin2\sqrt{x},v=cos2\sqrt{x}$
${W}_{1/2}=\left|\begin{array}{rr}u& v\\ {u}^{1/2}& {v}^{1/2}\end{array}\right|=\left|\begin{array}{rr}cos2\sqrt{x}& sin2\sqrt{x}\\ -sin2\sqrt{x}& cos2\sqrt{x}\end{array}\right|=1$
$\begin{array}{rl}X& =\underset{\alpha}{\int}\frac{-vR}{{W}_{\alpha}}d{x}^{\alpha}=\underset{1/2}{\int}\frac{-tan2\sqrt{x}cos2\sqrt{x}}{1}d{x}^{1/2}\\ & =\int -{x}^{-1/2}sin2\sqrt{x}dx=cos2\sqrt{x},\end{array}$

*and*
$Y=\underset{\alpha}{\int}\frac{uR}{{W}_{\alpha}}d{x}^{\alpha}=\underset{1/2}{\int}\frac{tan2\sqrt{x}sin2\sqrt{x}}{1}d{x}^{1/2}$
$=\int \frac{{x}^{-1/2}{sin}^{2}2\sqrt{x}}{cos2\sqrt{x}}dx$
$=\{sin2\sqrt{x}-log(sec2\sqrt{x}+tan2\sqrt{x})\},$

*hence, the complete solution is*
$\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y=asin2\sqrt{x}+bcos2\sqrt{x}-log(sec2\sqrt{x}+tan2\sqrt{x})cos2\sqrt{x}$

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