Internal problem ID [5832]
Internal file name [OUTPUT/5080_Sunday_June_05_2022_03_20_51_PM_73272133/index.tex]
Book: Ordinary differential equations and calculus of variations. Makarets and Reshetnyak. Wold
Scientific. Singapore. 1995
Section: Chapter 2. Linear homogeneous equations. Section 2.3.4 problems. page 104
Problem number: 11.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_non_constant_coeff_transformation_on_B"
Maple gives the following as the ode type
[[_2nd_order, _linear, _nonhomogeneous]]
\[ \boxed {\left (\cos \left (x \right )+\sin \left (x \right )\right ) y^{\prime \prime }-2 \cos \left (x \right ) y^{\prime }+\left (\cos \left (x \right )-\sin \left (x \right )\right ) y=\left (\cos \left (x \right )+\sin \left (x \right )\right )^{2} {\mathrm e}^{2 x}} \]
Given an ode of the form \begin {align*} A y^{\prime \prime } + B y^{\prime } + C y &= F(x) \end {align*}
This method reduces the order ode the ODE by one by applying the transformation \begin {align*} y&= B v \end {align*}
This results in \begin {align*} y' &=B' v+ v' B \\ y'' &=B'' v+ B' v' +v'' B + v' B' \\ &=v'' B+2 v'+ B'+B'' v \end {align*}
And now the original ode becomes \begin {align*} A\left ( v'' B+2v'B'+ B'' v\right )+B\left ( B'v+ v' B\right ) +CBv & =0\\ ABv'' +\left ( 2AB'+B^{2}\right ) v'+\left (AB''+BB'+CB\right ) v & =0 \tag {1} \end {align*}
If the term \(AB''+BB'+CB\) is zero, then this method works and can be used to solve \[ ABv''+\left ( 2AB' +B^{2}\right ) v'=0 \] By Using \(u=v'\) which reduces the order of the above ode to one. The new ode is \[ ABu'+\left ( 2AB'+B^{2}\right ) u=0 \] The above ode is first order ode which is solved for \(u\). Now a new ode \(v'=u\) is solved for \(v\) as first order ode. Then the final solution is obtain from \(y=Bv\).
This method works only if the term \(AB''+BB'+CB\) is zero. The given ODE shows that \begin {align*} A &= \cos \left (x \right )+\sin \left (x \right )\\ B &= -2 \cos \left (x \right )\\ C &= \cos \left (x \right )-\sin \left (x \right )\\ F &= {\mathrm e}^{2 x} \left (1+\sin \left (2 x \right )\right ) \end {align*}
The above shows that for this ode \begin {align*} AB''+BB'+CB &= \left (\cos \left (x \right )+\sin \left (x \right )\right ) \left (2 \cos \left (x \right )\right ) + \left (-2 \cos \left (x \right )\right ) \left (2 \sin \left (x \right )\right ) + \left (\cos \left (x \right )-\sin \left (x \right )\right ) \left (-2 \cos \left (x \right )\right ) \\ &=2 \left (\cos \left (x \right )+\sin \left (x \right )\right ) \cos \left (x \right )-4 \cos \left (x \right ) \sin \left (x \right )-2 \cos \left (x \right ) \left (\cos \left (x \right )-\sin \left (x \right )\right )\\ &=0 \end {align*}
Hence the ode in \(v\) given in (1) now simplifies to \begin {align*} -\cos \left (2 x \right )-1-\sin \left (2 x \right ) v'' +\left ( 4+2 \sin \left (2 x \right )\right ) v' & =0 \end {align*}
Now by applying \(v'=u\) the above becomes \begin {align*} -\left (\cos \left (2 x \right )+\sin \left (2 x \right )+1\right ) u^{\prime }\left (x \right )+2 \left (2+\sin \left (2 x \right )\right ) u \left (x \right ) = 0 \end {align*}
Which is now solved for \(u\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {2 \left (2+\sin \left (2 x \right )\right ) u}{\cos \left (2 x \right )+\sin \left (2 x \right )+1} \end {align*}
Where \(f(x)=\frac {4+2 \sin \left (2 x \right )}{\cos \left (2 x \right )+\sin \left (2 x \right )+1}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= \frac {4+2 \sin \left (2 x \right )}{\cos \left (2 x \right )+\sin \left (2 x \right )+1} \,d x\\ \int { \frac {1}{u} \,du} &= \int {\frac {4+2 \sin \left (2 x \right )}{\cos \left (2 x \right )+\sin \left (2 x \right )+1} \,d x}\\ \ln \left (u \right )&=\ln \left (\tan \left (x \right )+1\right )+\frac {\ln \left (1+\tan \left (x \right )^{2}\right )}{2}+x +c_{1}\\ u&={\mathrm e}^{\ln \left (\tan \left (x \right )+1\right )+\frac {\ln \left (1+\tan \left (x \right )^{2}\right )}{2}+x +c_{1}}\\ &=c_{1} {\mathrm e}^{\ln \left (\tan \left (x \right )+1\right )+\frac {\ln \left (1+\tan \left (x \right )^{2}\right )}{2}+x} \end {align*}
The ode for \(v\) now becomes \begin {align*} v' &= u\\ &=c_{1} {\mathrm e}^{\ln \left (\tan \left (x \right )+1\right )+\frac {\ln \left (1+\tan \left (x \right )^{2}\right )}{2}+x} \end {align*}
Which is now solved for \(v\). Integrating both sides gives \begin {align*} v \left (x \right ) &= \int { c_{1} {\mathrm e}^{\ln \left (\tan \left (x \right )+1\right )+\frac {\ln \left (1+\tan \left (x \right )^{2}\right )}{2}+x}\,\mathop {\mathrm {d}x}}\\ &= c_{1} {\mathrm e}^{\frac {\ln \left (1+\tan \left (x \right )^{2}\right )}{2}+x}+c_{2} \end {align*}
Therefore the homogeneous solution is \begin {align*} y_h(x) &= B v\\ &= \left (-2 \cos \left (x \right )\right ) \left (c_{1} {\mathrm e}^{\frac {\ln \left (1+\tan \left (x \right )^{2}\right )}{2}+x}+c_{2}\right ) \\ &= -2 c_{1} \operatorname {csgn}\left (\sec \left (x \right )\right ) {\mathrm e}^{x}-2 c_{2} \cos \left (x \right ) \end {align*}
And now the particular solution \(y_p(x)\) will be found. The particular solution \(y_p\) can be found using either the method of undetermined coefficients, or the method of variation of parameters. The method of variation of parameters will be used as it is more general and can be used when the coefficients of the ODE depend on \(x\) as well. Let \begin{equation} \tag{1} y_p(x) = u_1 y_1 + u_2 y_2 \end{equation} Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the two linearly independent solutions of the homogeneous ODE) found earlier when solving the homogeneous ODE as \begin{align*} y_1 &= {\mathrm e}^{x} \operatorname {csgn}\left (\sec \left (x \right )\right ) \\ y_2 &= \cos \left (x \right ) \\ \end{align*} In the Variation of parameters \(u_1,u_2\) are found using \begin{align*} \tag{2} u_1 &= -\int \frac {y_2 f(x)}{a W(x)} \\ \tag{3} u_2 &= \int \frac {y_1 f(x)}{a W(x)} \\ \end{align*} Where \(W(x)\) is the Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The Wronskian is given by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence \[ W = \begin {vmatrix} {\mathrm e}^{x} \operatorname {csgn}\left (\sec \left (x \right )\right ) & \cos \left (x \right ) \\ \frac {d}{dx}\left ({\mathrm e}^{x} \operatorname {csgn}\left (\sec \left (x \right )\right )\right ) & \frac {d}{dx}\left (\cos \left (x \right )\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} {\mathrm e}^{x} \operatorname {csgn}\left (\sec \left (x \right )\right ) & \cos \left (x \right ) \\ {\mathrm e}^{x} \operatorname {csgn}\left (\sec \left (x \right )\right )+{\mathrm e}^{x} \operatorname {csgn}\left (1, \sec \left (x \right )\right ) \sec \left (x \right ) \tan \left (x \right ) & -\sin \left (x \right ) \end {vmatrix} \] Therefore \[ W = \left ({\mathrm e}^{x} \operatorname {csgn}\left (\sec \left (x \right )\right )\right )\left (-\sin \left (x \right )\right ) - \left (\cos \left (x \right )\right )\left ({\mathrm e}^{x} \operatorname {csgn}\left (\sec \left (x \right )\right )+{\mathrm e}^{x} \operatorname {csgn}\left (1, \sec \left (x \right )\right ) \sec \left (x \right ) \tan \left (x \right )\right ) \] Which simplifies to \[ W = -\cos \left (x \right ) \operatorname {csgn}\left (1, \sec \left (x \right )\right ) \sec \left (x \right ) {\mathrm e}^{x} \tan \left (x \right )-\operatorname {csgn}\left (\sec \left (x \right )\right ) {\mathrm e}^{x} \sin \left (x \right )-\cos \left (x \right ) {\mathrm e}^{x} \operatorname {csgn}\left (\sec \left (x \right )\right ) \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\cos \left (x \right ) {\mathrm e}^{2 x} \left (1+\sin \left (2 x \right )\right )}{\left (\cos \left (x \right )+\sin \left (x \right )\right ) \left (-\cos \left (x \right ) \operatorname {csgn}\left (1, \sec \left (x \right )\right ) \sec \left (x \right ) {\mathrm e}^{x} \tan \left (x \right )-\operatorname {csgn}\left (\sec \left (x \right )\right ) {\mathrm e}^{x} \sin \left (x \right )-\cos \left (x \right ) {\mathrm e}^{x} \operatorname {csgn}\left (\sec \left (x \right )\right )\right )}\,dx \] Hence \[ u_1 = -\left (\int _{0}^{x}\frac {\cos \left (\alpha \right ) {\mathrm e}^{2 \alpha } \left (1+\sin \left (2 \alpha \right )\right )}{\left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right ) \left (-\cos \left (\alpha \right ) \operatorname {csgn}\left (1, \sec \left (\alpha \right )\right ) \sec \left (\alpha \right ) {\mathrm e}^{\alpha } \tan \left (\alpha \right )-\operatorname {csgn}\left (\sec \left (\alpha \right )\right ) {\mathrm e}^{\alpha } \sin \left (\alpha \right )-\cos \left (\alpha \right ) {\mathrm e}^{\alpha } \operatorname {csgn}\left (\sec \left (\alpha \right )\right )\right )}d \alpha \right ) \] And Eq. (3) becomes \[ u_2 = \int \frac {{\mathrm e}^{x} \operatorname {csgn}\left (\sec \left (x \right )\right ) {\mathrm e}^{2 x} \left (1+\sin \left (2 x \right )\right )}{\left (\cos \left (x \right )+\sin \left (x \right )\right ) \left (-\cos \left (x \right ) \operatorname {csgn}\left (1, \sec \left (x \right )\right ) \sec \left (x \right ) {\mathrm e}^{x} \tan \left (x \right )-\operatorname {csgn}\left (\sec \left (x \right )\right ) {\mathrm e}^{x} \sin \left (x \right )-\cos \left (x \right ) {\mathrm e}^{x} \operatorname {csgn}\left (\sec \left (x \right )\right )\right )}\,dx \] Hence \[ u_2 = \int _{0}^{x}\frac {{\mathrm e}^{\alpha } \operatorname {csgn}\left (\sec \left (\alpha \right )\right ) {\mathrm e}^{2 \alpha } \left (1+\sin \left (2 \alpha \right )\right )}{\left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right ) \left (-\cos \left (\alpha \right ) \operatorname {csgn}\left (1, \sec \left (\alpha \right )\right ) \sec \left (\alpha \right ) {\mathrm e}^{\alpha } \tan \left (\alpha \right )-\operatorname {csgn}\left (\sec \left (\alpha \right )\right ) {\mathrm e}^{\alpha } \sin \left (\alpha \right )-\cos \left (\alpha \right ) {\mathrm e}^{\alpha } \operatorname {csgn}\left (\sec \left (\alpha \right )\right )\right )}d \alpha \] Which simplifies to \begin{align*} u_1 &= \int _{0}^{x}\frac {{\mathrm e}^{\alpha } \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right ) \cos \left (\alpha \right )^{2}}{\operatorname {csgn}\left (1, \sec \left (\alpha \right )\right ) \sin \left (\alpha \right )+\cos \left (\alpha \right ) \operatorname {csgn}\left (\sec \left (\alpha \right )\right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}d \alpha \\ u_2 &= -\left (\int _{0}^{x}\frac {{\mathrm e}^{2 \alpha } \operatorname {csgn}\left (\sec \left (\alpha \right )\right ) \cos \left (\alpha \right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}{\operatorname {csgn}\left (1, \sec \left (\alpha \right )\right ) \sin \left (\alpha \right )+\operatorname {csgn}\left (\sec \left (\alpha \right )\right ) \cos \left (\alpha \right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}d \alpha \right ) \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = \left (\int _{0}^{x}\frac {{\mathrm e}^{\alpha } \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right ) \cos \left (\alpha \right )^{2}}{\operatorname {csgn}\left (1, \sec \left (\alpha \right )\right ) \sin \left (\alpha \right )+\cos \left (\alpha \right ) \operatorname {csgn}\left (\sec \left (\alpha \right )\right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}d \alpha \right ) {\mathrm e}^{x} \operatorname {csgn}\left (\sec \left (x \right )\right )-\left (\int _{0}^{x}\frac {{\mathrm e}^{2 \alpha } \operatorname {csgn}\left (\sec \left (\alpha \right )\right ) \cos \left (\alpha \right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}{\operatorname {csgn}\left (1, \sec \left (\alpha \right )\right ) \sin \left (\alpha \right )+\operatorname {csgn}\left (\sec \left (\alpha \right )\right ) \cos \left (\alpha \right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}d \alpha \right ) \cos \left (x \right ) \] Hence the complete solution is \begin {align*} y(x) &= y_h + y_p \\ &= \left (-2 c_{1} \operatorname {csgn}\left (\sec \left (x \right )\right ) {\mathrm e}^{x}-2 c_{2} \cos \left (x \right )\right ) + \left (\left (\int _{0}^{x}\frac {{\mathrm e}^{\alpha } \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right ) \cos \left (\alpha \right )^{2}}{\operatorname {csgn}\left (1, \sec \left (\alpha \right )\right ) \sin \left (\alpha \right )+\cos \left (\alpha \right ) \operatorname {csgn}\left (\sec \left (\alpha \right )\right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}d \alpha \right ) {\mathrm e}^{x} \operatorname {csgn}\left (\sec \left (x \right )\right )-\left (\int _{0}^{x}\frac {{\mathrm e}^{2 \alpha } \operatorname {csgn}\left (\sec \left (\alpha \right )\right ) \cos \left (\alpha \right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}{\operatorname {csgn}\left (1, \sec \left (\alpha \right )\right ) \sin \left (\alpha \right )+\operatorname {csgn}\left (\sec \left (\alpha \right )\right ) \cos \left (\alpha \right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}d \alpha \right ) \cos \left (x \right )\right )\\ &= -2 c_{1} \operatorname {csgn}\left (\sec \left (x \right )\right ) {\mathrm e}^{x}-2 c_{2} \cos \left (x \right )+\left (\int _{0}^{x}\frac {{\mathrm e}^{\alpha } \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right ) \cos \left (\alpha \right )^{2}}{\operatorname {csgn}\left (1, \sec \left (\alpha \right )\right ) \sin \left (\alpha \right )+\cos \left (\alpha \right ) \operatorname {csgn}\left (\sec \left (\alpha \right )\right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}d \alpha \right ) {\mathrm e}^{x} \operatorname {csgn}\left (\sec \left (x \right )\right )-\left (\int _{0}^{x}\frac {{\mathrm e}^{2 \alpha } \operatorname {csgn}\left (\sec \left (\alpha \right )\right ) \cos \left (\alpha \right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}{\operatorname {csgn}\left (1, \sec \left (\alpha \right )\right ) \sin \left (\alpha \right )+\operatorname {csgn}\left (\sec \left (\alpha \right )\right ) \cos \left (\alpha \right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}d \alpha \right ) \cos \left (x \right ) \end {align*}
Simplifying the solution \(y = -2 c_{1} \operatorname {csgn}\left (\sec \left (x \right )\right ) {\mathrm e}^{x}-2 c_{2} \cos \left (x \right )+\left (\int _{0}^{x}\frac {{\mathrm e}^{\alpha } \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right ) \cos \left (\alpha \right )^{2}}{\operatorname {csgn}\left (1, \sec \left (\alpha \right )\right ) \sin \left (\alpha \right )+\cos \left (\alpha \right ) \operatorname {csgn}\left (\sec \left (\alpha \right )\right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}d \alpha \right ) {\mathrm e}^{x} \operatorname {csgn}\left (\sec \left (x \right )\right )-\left (\int _{0}^{x}\frac {{\mathrm e}^{2 \alpha } \operatorname {csgn}\left (\sec \left (\alpha \right )\right ) \cos \left (\alpha \right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}{\operatorname {csgn}\left (1, \sec \left (\alpha \right )\right ) \sin \left (\alpha \right )+\operatorname {csgn}\left (\sec \left (\alpha \right )\right ) \cos \left (\alpha \right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}d \alpha \right ) \cos \left (x \right )\) to \(y = -2 c_{1} {\mathrm e}^{x}-2 c_{2} \cos \left (x \right )+\left (\int _{0}^{x}\frac {{\mathrm e}^{\alpha } \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right ) \cos \left (\alpha \right )^{2}}{\sin \left (\alpha \right )+\cos \left (\alpha \right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}d \alpha \right ) {\mathrm e}^{x}-\left (\int _{0}^{x}\frac {{\mathrm e}^{2 \alpha } \cos \left (\alpha \right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}{\sin \left (\alpha \right )+\left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right ) \cos \left (\alpha \right )}d \alpha \right ) \cos \left (x \right )\)
The solution(s) found are the following \begin{align*} \tag{1} y &= -2 c_{1} {\mathrm e}^{x}-2 c_{2} \cos \left (x \right )+\left (\int _{0}^{x}\frac {{\mathrm e}^{\alpha } \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right ) \cos \left (\alpha \right )^{2}}{\sin \left (\alpha \right )+\cos \left (\alpha \right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}d \alpha \right ) {\mathrm e}^{x}-\left (\int _{0}^{x}\frac {{\mathrm e}^{2 \alpha } \cos \left (\alpha \right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}{\sin \left (\alpha \right )+\left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right ) \cos \left (\alpha \right )}d \alpha \right ) \cos \left (x \right ) \\ \end{align*}
Verification of solutions
\[ y = -2 c_{1} {\mathrm e}^{x}-2 c_{2} \cos \left (x \right )+\left (\int _{0}^{x}\frac {{\mathrm e}^{\alpha } \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right ) \cos \left (\alpha \right )^{2}}{\sin \left (\alpha \right )+\cos \left (\alpha \right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}d \alpha \right ) {\mathrm e}^{x}-\left (\int _{0}^{x}\frac {{\mathrm e}^{2 \alpha } \cos \left (\alpha \right ) \left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right )}{\sin \left (\alpha \right )+\left (\cos \left (\alpha \right )+\sin \left (\alpha \right )\right ) \cos \left (\alpha \right )}d \alpha \right ) \cos \left (x \right ) \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 2; linear nonhomogeneous with symmetry [0,1] trying a double symmetry of the form [xi=0, eta=F(x)] trying symmetries linear in x and y(x) -> Try solving first the homogeneous part of the ODE trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) <- linear symmetries successful Change of variables used: [x = arccos(t)] Linear ODE actually solved: (t-(-t^2+1)^(1/2))*u(t)+((-t^2+1)^(1/2)*t-t^2)*diff(u(t),t)+(-(-t^2+1)^(1/2)*t^2-t^3+(-t^2+1)^(1/2)+t)*diff(diff(u(t),t),t) <- change of variables successful <- solving first the homogeneous part of the ODE successful`
✓ Solution by Maple
Time used: 0.422 (sec). Leaf size: 322
dsolve((cos(x)+sin(x))*diff(y(x),x$2)-2*cos(x)*diff(y(x),x)+(cos(x)-sin(x))*y(x)=(cos(x)+sin(x))^2*exp(2*x),y(x), singsol=all)
\[ y \left (x \right ) = -\cos \left (x \right ) \left (\left (\int {\mathrm e}^{\int \frac {\left (-\cot \left (x \right )+1\right ) \cos \left (x \right )+2 \sin \left (x \right ) \left (\tan \left (x \right )+1\right )}{\cos \left (x \right )+\sin \left (x \right )}d x} \sin \left (x \right )d x \right ) c_{1} -\left (\int {\mathrm e}^{2 x -2 \left (\int \frac {\sin \left (x \right )}{\cos \left (x \right )+\sin \left (x \right )}d x \right )-2 \left (\int \frac {\sin \left (x \right ) \tan \left (x \right )}{\cos \left (x \right )+\sin \left (x \right )}d x \right )+\int \frac {\cos \left (x \right ) \cot \left (x \right )}{\cos \left (x \right )+\sin \left (x \right )}d x -\left (\int \frac {\cos \left (x \right )}{\cos \left (x \right )+\sin \left (x \right )}d x \right )} \left (\csc \left (x \right )+\sec \left (x \right )\right )d x \right ) \left (\int {\mathrm e}^{2 \left (\int \frac {\sin \left (x \right )}{\cos \left (x \right )+\sin \left (x \right )}d x \right )+2 \left (\int \frac {\sin \left (x \right ) \tan \left (x \right )}{\cos \left (x \right )+\sin \left (x \right )}d x \right )-\left (\int \frac {\cos \left (x \right ) \cot \left (x \right )}{\cos \left (x \right )+\sin \left (x \right )}d x \right )+\int \frac {\cos \left (x \right )}{\cos \left (x \right )+\sin \left (x \right )}d x} \sin \left (x \right )d x \right )+\int {\mathrm e}^{2 x -2 \left (\int \frac {\sin \left (x \right )}{\cos \left (x \right )+\sin \left (x \right )}d x \right )-2 \left (\int \frac {\sin \left (x \right ) \tan \left (x \right )}{\cos \left (x \right )+\sin \left (x \right )}d x \right )+\int \frac {\cos \left (x \right ) \cot \left (x \right )}{\cos \left (x \right )+\sin \left (x \right )}d x -\left (\int \frac {\cos \left (x \right )}{\cos \left (x \right )+\sin \left (x \right )}d x \right )} \left (\csc \left (x \right )+\sec \left (x \right )\right ) \left (\int {\mathrm e}^{2 \left (\int \frac {\sin \left (x \right )}{\cos \left (x \right )+\sin \left (x \right )}d x \right )+2 \left (\int \frac {\sin \left (x \right ) \tan \left (x \right )}{\cos \left (x \right )+\sin \left (x \right )}d x \right )-\left (\int \frac {\cos \left (x \right ) \cot \left (x \right )}{\cos \left (x \right )+\sin \left (x \right )}d x \right )+\int \frac {\cos \left (x \right )}{\cos \left (x \right )+\sin \left (x \right )}d x} \sin \left (x \right )d x \right )d x -c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 4.817 (sec). Leaf size: 476
DSolve[(Cos[x]+Sin[x])*y''[x]-2*Cos[x]*y'[x]+(Cos[x]-Sin[x])*y[x]==(Cos[x]+Sin[x])^2*Exp[2*x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \left (e^{-2 i x}\right )^{\frac {1}{2}-\frac {i}{2}} \left (e^{i x}\right )^{1-2 i} \left (-\frac {i \left (-1+e^{2 i \arctan \left (e^{-2 i x}\right )}\right )}{1+e^{2 i \arctan \left (e^{-2 i x}\right )}}\right )^{-\frac {1}{2}-\frac {i}{2}} \left (-i \left (e^{-2 i x}\right )^i \sqrt {1+e^{-4 i x}} \sqrt {1+e^{4 i x}} e^{2 i \left (2 x+\arctan \left (e^{-2 i x}\right )\right )}-2 i \sqrt {-e^{4 i x}} \sqrt {-\left (1+e^{4 i x}\right )^2} e^{2 i \arctan \left (e^{-2 i x}\right )} \left (-\frac {i \left (-1+e^{2 i \arctan \left (e^{-2 i x}\right )}\right )}{1+e^{2 i \arctan \left (e^{-2 i x}\right )}}\right )^i+e^{4 i x} \left (e^{-2 i x}\right )^i \sqrt {1+e^{-4 i x}} \sqrt {1+e^{4 i x}}\right )}{\sqrt {-e^{4 i x}} \sqrt {-\left (1+e^{4 i x}\right )^2} \left (1+e^{2 i \arctan \left (e^{-2 i x}\right )}\right )}+\frac {c_2 e^{3 i x} \left (e^{-2 i x}\right )^{\frac {1}{2}+\frac {i}{2}} \sqrt {1+e^{-4 i x}} \left (e^{2 i \arctan \left (e^{-2 i x}\right )}+i\right ) \left (-\frac {i \left (-1+e^{2 i \arctan \left (e^{-2 i x}\right )}\right )}{1+e^{2 i \arctan \left (e^{-2 i x}\right )}}\right )^{\frac {1}{2}-\frac {i}{2}}}{\sqrt {1+e^{4 i x}} \left (-1+e^{2 i \arctan \left (e^{-2 i x}\right )}\right )}+c_1 \left (e^{i x}\right )^{-i} \]