2.3.6 Problem 6

Internal problem ID [8864]
Book : Own collection of miscellaneous problems
Section : section 3.0
Problem number : 6
Date solved : Wednesday, March 05, 2025 at 06:57:01 AM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

Solve

$$\begin{array}{rl}{y}^{\prime \prime }+y& =\sin \left(x\right)\end{array}$$

With initial conditions

$$\begin{array}{rl}y\left(1\right)& =0\end{array}$$

Solved as second order linear constant coeff ode

Time used: 0.250 (sec)

This is second order non-homogeneous ODE. In standard form the ODE is

$$A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=f(x)$$

Where $A=1,B=0,C=1,f(x)=\sin \left(x\right)$. Let the solution be

$$y=y_h+y_p$$

Where $y_h$ is the solution to the homogeneous ODE $A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=0$, and $y_p$ is a particular solution to the non-homogeneous ODE $A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=f(x)$. $y_h$ is the solution to

$$y^{\prime \prime }+y=0$$

This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

$$A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=0$$

Where in the above $A=1,B=0,C=1$. Let the solution be $y=e^{\lambda x}$. Substituting this into the ODE gives

$$\begin{array}{}\text{(1)}& {\lambda }^2{\mathrm{e}}^{x\lambda }+{\mathrm{e}}^{x\lambda }=0\end{array}$$

Since exponential function is never zero, then dividing Eq(2) throughout by $e^{\lambda x}$ gives

$$\begin{array}{}\text{(2)}& {\lambda }^2+1=0\end{array}$$

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula

$${\lambda }_{1,2}=\frac{-B}{2A}\pm \frac{1}{2A}\sqrt{B^2-4AC}$$

Substituting $A=1,B=0,C=1$ into the above gives

$$\begin{array}{rl}{\lambda }_{1,2}& =\frac{0}{(2)\left(1\right)}\pm \frac{1}{(2)\left(1\right)}\sqrt{0^2-(4)\left(1\right)\left(1\right)}\\ & =\pm i\end{array}$$

Hence

$$\begin{array}{rl}{\lambda }_1& =+i\\ {\lambda }_2& =-i\end{array}$$

Which simplifies to

$$\begin{array}{rl}{\lambda }_1& =i\\ {\lambda }_2& =-i\end{array}$$

Since roots are complex conjugate of each others, then let the roots be

$${\lambda }_{1,2}=\alpha \pm i\beta$$

Where $\alpha =0$ and $\beta =1$. Therefore the final solution, when using Euler relation, can be written as

$$y=e^{\alpha x}(c_1\cos (\beta x)+c_2\sin (\beta x))$$

Which becomes

$$y=e^0(c_1\cos \left(x\right)+c_2\sin \left(x\right))$$

Or

$$y=c_1\cos \left(x\right)+c_2\sin \left(x\right)$$

Therefore the homogeneous solution $y_h$ is

$$y_h=c_1\cos \left(x\right)+c_2\sin \left(x\right)$$

The particular solution is now found using the method of undetermined coefficients. Looking at the RHS of the ode, which is

$$\sin \left(x\right)$$

Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is

$$[\{\cos \left(x\right),\sin \left(x\right)\}]$$

While the set of the basis functions for the homogeneous solution found earlier is

$$\{\cos \left(x\right),\sin \left(x\right)\}$$

Since $\cos \left(x\right)$ is duplicated in the UC_set, then this basis is multiplied by extra $x$. The UC_set becomes

$$[\{x\cos \left(x\right),x\sin \left(x\right)\}]$$

Since there was duplication between the basis functions in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis function in the above updated UC_set.

$$y_p=A_{1}x\cos \left(x\right)+A_{2}x\sin \left(x\right)$$

The unknowns $\{A_1,A_2\}$ are found by substituting the above trial solution $y_p$ into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives

$$-2A_1\sin \left(x\right)+2A_2\cos \left(x\right)=\sin \left(x\right)$$

Solving for the unknowns by comparing coefficients results in

$$[A_1=-\frac{1}{2},A_2=0]$$

Substituting the above back in the above trial solution $y_p$, gives the particular solution

$$y_p=-\frac{x\cos \left(x\right)}{2}$$

Therefore the general solution is

$$\begin{array}{rl}y& =y_h+y_p\\ & =(c_1\cos \left(x\right)+c_2\sin \left(x\right))+(-\frac{x\cos \left(x\right)}{2})\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =(-c_2\tan \left(1\right)+\frac{1}{2})\cos \left(x\right)+c_2\sin \left(x\right)-\frac{x\cos \left(x\right)}{2}\end{array}$$

Figure 2.185: Slope field $y^{\prime \prime }+y=\sin \left(x\right)$

Solved as second order ode using Kovacic algorithm

Time used: 0.185 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& y^{\prime \prime }+y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =1\\ \text{(3)}& B& =0C& =1\end{array}$$

Applying the Liouville transformation on the dependent variable gives

$$\begin{array}{rl}z(x)& =y{e}^{\int \frac{B}{2A}dx}\end{array}$$

Then (2) becomes

$$\begin{array}{}\text{(4)}& z^{″}(x)=rz(x)\end{array}$$

Where $r$ is given by

$$\begin{array}{rl}\text{(5)}& r& =\frac{s}{t}& =\frac{2A{B}^{\prime }-2B{A}^{\prime }+B^2-4AC}{4A^2}\end{array}$$

Substituting the values of $A,B,C$ from (3) in the above and simplifying gives

$$\begin{array}{}\text{(6)}& r& =\frac{-1}{1}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-1\\ t& =1\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =-z\left(x\right)\end{array}$$

Equation (7) is now solved. After finding $z(x)$ then $y$ is found using the inverse transformation

$$\begin{array}{rl}y& =z\left(x\right)e^{-\int \frac{B}{2A}dx}\end{array}$$

The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of $r$ and the order of $r$ at $\mathrm{\infty }$. The following table summarizes these cases.

Case	Allowed pole order for $r$	Allowed value for $\mathcal{O}(\mathrm{\infty })$
1	$\{0,1,2,4,6,8,\cdots \}$	$\{\cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \}$
2	Need to have at least one pole that is either order $2$ or odd order greater than $2$. Any other pole order is allowed as long as the above condition is satisfied. Hence the following set of pole orders are all allowed. $\{1,2\}$,$\{1,3\}$,$\{2\}$,$\{3\}$,$\{3,4\}$,$\{1,2,5\}$.	no condition
3	$\{1,2\}$	$\{2,3,4,5,6,7,\cdots \}$

Table 2.41: Necessary conditions for each Kovacic case

The order of $r$ at $\mathrm{\infty }$ is the degree of $t$ minus the degree of $s$. Therefore

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =0-0\\ & =0\end{array}$$

There are no poles in $r$. Therefore the set of poles $\mathrm{\Gamma }$ is empty. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $0$ then the necessary conditions for case one are met. Therefore

$$\begin{array}{rl}L& =[1]\end{array}$$

Since $r=-1$ is not a function of $x$, then there is no need run Kovacic algorithm to obtain a solution for transformed ode $z^{″}=rz$ as one solution is

$$z_1(x)=\cos \left(x\right)$$

Using the above, the solution for the original ode can now be found. The first solution to the original ode in $y$ is found from

$$y_1=z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}$$

Since $B=0$ then the above reduces to

$$\begin{array}{rl}{y}_1& =z_1\\ & =\cos \left(x\right)\end{array}$$

Which simplifies to

$$y_1=\cos \left(x\right)$$

The second solution $y_2$ to the original ode is found using reduction of order

$$y_2=y_1\int \frac{e^{\int -\frac{B}{A}dx}}{y_1^2}dx$$

Since $B=0$ then the above becomes

$$\begin{array}{rl}{y}_2& =y_1\int \frac{1}{y_1^2}dx\\ & =\cos \left(x\right)\int \frac{1}{\cos {\left(x\right)}^2}dx\\ & =\cos \left(x\right)(\tan \left(x\right))\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1(\cos \left(x\right))+c_2(\cos \left(x\right)(\tan \left(x\right)))\end{array}$$

This is second order nonhomogeneous ODE. Let the solution be

$$y=y_h+y_p$$

Where $y_h$ is the solution to the homogeneous ODE $A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=0$, and $y_p$ is a particular solution to the nonhomogeneous ODE $A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=f(x)$. $y_h$ is the solution to

$$y^{\prime \prime }+y=0$$

The homogeneous solution is found using the Kovacic algorithm which results in

$$y_h=c_1\cos \left(x\right)+c_2\sin \left(x\right)$$

The particular solution is now found using the method of undetermined coefficients. Looking at the RHS of the ode, which is

$$\sin \left(x\right)$$

Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is

$$[\{\cos \left(x\right),\sin \left(x\right)\}]$$

While the set of the basis functions for the homogeneous solution found earlier is

$$\{\cos \left(x\right),\sin \left(x\right)\}$$

Since $\cos \left(x\right)$ is duplicated in the UC_set, then this basis is multiplied by extra $x$. The UC_set becomes

$$[\{x\cos \left(x\right),x\sin \left(x\right)\}]$$

Since there was duplication between the basis functions in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis function in the above updated UC_set.

$$y_p=A_{1}x\cos \left(x\right)+A_{2}x\sin \left(x\right)$$

The unknowns $\{A_1,A_2\}$ are found by substituting the above trial solution $y_p$ into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives

$$-2A_1\sin \left(x\right)+2A_2\cos \left(x\right)=\sin \left(x\right)$$

Solving for the unknowns by comparing coefficients results in

$$[A_1=-\frac{1}{2},A_2=0]$$

Substituting the above back in the above trial solution $y_p$, gives the particular solution

$$y_p=-\frac{x\cos \left(x\right)}{2}$$

Therefore the general solution is

$$\begin{array}{rl}y& =y_h+y_p\\ & =(c_1\cos \left(x\right)+c_2\sin \left(x\right))+(-\frac{x\cos \left(x\right)}{2})\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =(-c_2\tan \left(1\right)+\frac{1}{2})\cos \left(x\right)+c_2\sin \left(x\right)-\frac{x\cos \left(x\right)}{2}\end{array}$$

Figure 2.186: Slope field $y^{\prime \prime }+y=\sin \left(x\right)$

Solved as second order ode adjoint method

Time used: 1.908 (sec)

In normal form the ode

$$\begin{array}{}\text{(1)}& y^{\prime \prime }+y=\sin \left(x\right)\end{array}$$

Becomes

$$\begin{array}{}\text{(2)}& y^{\prime \prime }+p\left(x\right)y^{\prime }+q\left(x\right)y& =r\left(x\right)\end{array}$$

Where

$$\begin{array}{rl}p\left(x\right)& =0\\ q\left(x\right)& =1\\ r\left(x\right)& =\sin \left(x\right)\end{array}$$

The Lagrange adjoint ode is given by

$$\begin{array}{rl}{\xi }^{{}^{″}}-(\xi p{)}^{\prime }+\xi q& =0\\ {\xi }^{{}^{″}}-{\left(0\right)}^{\prime }+\left(\xi \left(x\right)\right)& =0\\ {\xi }^{\prime \prime }\left(x\right)+\xi \left(x\right)& =0\end{array}$$

Which is solved for $\xi (x)$. This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

$$A{\xi }^{″}(x)+B{\xi }^{\prime }(x)+C\xi (x)=0$$

Where in the above $A=1,B=0,C=1$. Let the solution be $\xi =e^{\lambda x}$. Substituting this into the ODE gives

$$\begin{array}{}\text{(1)}& {\lambda }^2{\mathrm{e}}^{x\lambda }+{\mathrm{e}}^{x\lambda }=0\end{array}$$

Since exponential function is never zero, then dividing Eq(2) throughout by $e^{\lambda x}$ gives

$$\begin{array}{}\text{(2)}& {\lambda }^2+1=0\end{array}$$

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula

$${\lambda }_{1,2}=\frac{-B}{2A}\pm \frac{1}{2A}\sqrt{B^2-4AC}$$

Substituting $A=1,B=0,C=1$ into the above gives

$$\begin{array}{rl}{\lambda }_{1,2}& =\frac{0}{(2)\left(1\right)}\pm \frac{1}{(2)\left(1\right)}\sqrt{0^2-(4)\left(1\right)\left(1\right)}\\ & =\pm i\end{array}$$

Hence

$$\begin{array}{rl}{\lambda }_1& =+i\\ {\lambda }_2& =-i\end{array}$$

Which simplifies to

$$\begin{array}{rl}{\lambda }_1& =i\\ {\lambda }_2& =-i\end{array}$$

Since roots are complex conjugate of each others, then let the roots be

$${\lambda }_{1,2}=\alpha \pm i\beta$$

Where $\alpha =0$ and $\beta =1$. Therefore the final solution, when using Euler relation, can be written as

$$\xi =e^{\alpha x}(c_1\cos (\beta x)+c_2\sin (\beta x))$$

Which becomes

$$\xi =e^0(c_1\cos \left(x\right)+c_2\sin \left(x\right))$$

Or

$$\xi =c_1\cos \left(x\right)+c_2\sin \left(x\right)$$

Will add steps showing solving for IC soon.

The original ode now reduces to first order ode

$$\begin{array}{rl}\xi \left(x\right)y^{\prime }-y{\xi }^{\prime }\left(x\right)+\xi \left(x\right)p\left(x\right)y& =\int \xi \left(x\right)r\left(x\right)dx\\ y^{\prime }+y(p\left(x\right)-\frac{{\xi }^{\prime }\left(x\right)}{\xi \left(x\right)})& =\frac{\int \xi \left(x\right)r\left(x\right)dx}{\xi \left(x\right)}\end{array}$$

Or

$$\begin{array}{rl}{y}^{\prime }-\frac{y(-c_1\sin \left(x\right)+c_2\cos \left(x\right))}{c_1\cos \left(x\right)+c_2\sin \left(x\right)}& =\frac{-\frac{\cos {\left(x\right)}^2{c}_1}{2}+c_2(-\frac{\cos \left(x\right)\sin \left(x\right)}{2}+\frac{x}{2})}{c_1\cos \left(x\right)+c_2\sin \left(x\right)}\end{array}$$

Which is now a first order ode. This is now solved for $y$. In canonical form a linear first order is

$$\begin{array}{rl}{y}^{\prime }+q(x)y& =p(x)\end{array}$$

Comparing the above to the given ode shows that

$$\begin{array}{rl}q(x)& =-\frac{-c_1\sin \left(x\right)+c_2\cos \left(x\right)}{c_1\cos \left(x\right)+c_2\sin \left(x\right)}\\ p(x)& =\frac{-c_2\cos \left(x\right)\sin \left(x\right)-\cos {\left(x\right)}^2{c}_1+c_{2}x}{2c_2\sin \left(x\right)+2c_1\cos \left(x\right)}\end{array}$$

The integrating factor $\mu$ is

$$\begin{array}{rl}\mu & =e^{\int qdx}\\ & ={\mathrm{e}}^{\int -\frac{-c_1\sin \left(x\right)+c_2\cos \left(x\right)}{c_1\cos \left(x\right)+c_2\sin \left(x\right)}dx}\\ & =\frac{1}{c_1\cos \left(x\right)+c_2\sin \left(x\right)}\end{array}$$

The ode becomes

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}\left(\mu y\right)& =\mu p\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(\mu y\right)& =\left(\mu \right)\left(\frac{-c_2\cos \left(x\right)\sin \left(x\right)-\cos {\left(x\right)}^2{c}_1+c_{2}x}{2c_2\sin \left(x\right)+2c_1\cos \left(x\right)}\right)\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{y}{c_1\cos \left(x\right)+c_2\sin \left(x\right)}\right)& =\left(\frac{1}{c_1\cos \left(x\right)+c_2\sin \left(x\right)}\right)\left(\frac{-c_2\cos \left(x\right)\sin \left(x\right)-\cos {\left(x\right)}^2{c}_1+c_{2}x}{2c_2\sin \left(x\right)+2c_1\cos \left(x\right)}\right)\\ \mathrm{d}\left(\frac{y}{c_1\cos \left(x\right)+c_2\sin \left(x\right)}\right)& =\left(\frac{-c_2\cos \left(x\right)\sin \left(x\right)-\cos {\left(x\right)}^2{c}_1+c_{2}x}{(c_1\cos \left(x\right)+c_2\sin \left(x\right))(2c_2\sin \left(x\right)+2c_1\cos \left(x\right))}\right)\mathrm{d}x\end{array}$$

Integrating gives

$$\begin{array}{rl}\frac{y}{c_1\cos \left(x\right)+c_2\sin \left(x\right)}& =\int \frac{-c_2\cos \left(x\right)\sin \left(x\right)-\cos {\left(x\right)}^2{c}_1+c_{2}x}{(c_1\cos \left(x\right)+c_2\sin \left(x\right))(2c_2\sin \left(x\right)+2c_1\cos \left(x\right))}dx\\ & =-\frac{\cos \left(x\right)x}{2c_2\sin \left(x\right)+2c_1\cos \left(x\right)}+c_3\end{array}$$

Dividing throughout by the integrating factor $\frac{1}{c_1\cos \left(x\right)+c_2\sin \left(x\right)}$ gives the final solution

$$y=\frac{(2c_1{c}_3-x)\cos \left(x\right)}{2}+\sin \left(x\right)c_2{c}_3$$

Hence, the solution found using Lagrange adjoint equation method is

$$\begin{array}{rl}y& =\frac{(2c_1{c}_3-x)\cos \left(x\right)}{2}+\sin \left(x\right)c_2{c}_3\end{array}$$

The constants can be merged to give

$$y=\frac{(2c_1-x)\cos \left(x\right)}{2}+c_2\sin \left(x\right)$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =\frac{(-2c_2\tan \left(1\right)+1-x)\cos \left(x\right)}{2}+c_2\sin \left(x\right)\end{array}$$

Figure 2.187: Slope field $y^{\prime \prime }+y=\sin \left(x\right)$

✓ Maple. Time used: 0.072 (sec). Leaf size: 31

ode:=diff(diff(y(x),x),x)+y(x) = sin(x); 
ic:=y(1) = 0; 
dsolve([ode,ic],y(x), singsol=all);
 

$$y=\frac{((-2c_2-1)\tan \left(1\right)-x+1)\cos \left(x\right)}{2}+\frac{\sin \left(x\right)(2c_2+1)}{2}$$

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)] 
-> Try solving first the homogeneous part of the ODE 
   checking if the LODE has constant coefficients 
   <- constant coefficients successful 
<- solving first the homogeneous part of the ODE successful`
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & [\frac{d^2}{d{x}^2}y\left(x\right)+y\left(x\right)=\sin \left(x\right),y\left(1\right)=0]\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d^2}{d{x}^2}y\left(x\right)\\ \bullet & & \text{Characteristic polynomial of homogeneous ODE}\\ & & r^2+1=0\\ \bullet & & \text{Use quadratic formula to solve for}r\\ & & r=\frac{0\pm \left(\sqrt{-4}\right)}{2}\\ \bullet & & \text{Roots of the characteristic polynomial}\\ & & r=(-\mathrm{I},\mathrm{I})\\ \bullet & & \text{1st solution of the homogeneous ODE}\\ & & y_1\left(x\right)=\cos \left(x\right)\\ \bullet & & \text{2nd solution of the homogeneous ODE}\\ & & y_2\left(x\right)=\sin \left(x\right)\\ \bullet & & \text{General solution of the ODE}\\ & & y\left(x\right)=\mathit{C}\mathit{1}{y}_1\left(x\right)+\mathit{C}\mathit{2}{y}_2\left(x\right)+y_p\left(x\right)\\ \bullet & & \text{Substitute in solutions of the homogeneous ODE}\\ & & y\left(x\right)=\mathit{C}\mathit{1}\cos \left(x\right)+\mathit{C}\mathit{2}\sin \left(x\right)+y_p\left(x\right)\\ ◻& & \text{Find a particular solution}{y}_p\left(x\right)\text{of the ODE}\\ & \circ & \text{Use variation of parameters to find}{y}_p\text{here}f\left(x\right)\text{is the forcing function}\\ & & [y_p\left(x\right)=-y_1\left(x\right)(\int \frac{y_2\left(x\right)f\left(x\right)}{W(y_1\left(x\right),y_2\left(x\right))}dx)+y_2\left(x\right)(\int \frac{y_1\left(x\right)f\left(x\right)}{W(y_1\left(x\right),y_2\left(x\right))}dx),f\left(x\right)=\sin \left(x\right)]\\ & \circ & \text{Wronskian of solutions of the homogeneous equation}\\ & & W(y_1\left(x\right),y_2\left(x\right))=\left[\begin{array}{cc}\cos \left(x\right)& \sin \left(x\right)\\ -\sin \left(x\right)& \cos \left(x\right)\end{array}\right]\\ & \circ & \text{Compute Wronskian}\\ & & W(y_1\left(x\right),y_2\left(x\right))=1\\ & \circ & \text{Substitute functions into equation for}{y}_p\left(x\right)\\ & & y_p\left(x\right)=-\cos \left(x\right)(\int \sin {\left(x\right)}^{2}dx)+\frac{\sin \left(x\right)(\int \sin \left(2x\right)dx)}{2}\\ & \circ & \text{Compute integrals}\\ & & y_p\left(x\right)=\frac{\sin \left(x\right)}{4}-\frac{x\cos \left(x\right)}{2}\\ \bullet & & \text{Substitute particular solution into general solution to ODE}\\ & & y\left(x\right)=\mathit{C}\mathit{1}\cos \left(x\right)+\mathit{C}\mathit{2}\sin \left(x\right)+\frac{\sin \left(x\right)}{4}-\frac{x\cos \left(x\right)}{2}\end{array}$$

✓ Mathematica. Time used: 0.023 (sec). Leaf size: 18

ode=D[y[x],{x,2}]+y[x]==Sin[x]; 
ic={y[0] == 0}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to -\frac{1}{2}x\cos (x)+c_2\sin (x)$$

✓ Sympy. Time used: 0.103 (sec). Leaf size: 26

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(y(x) - sin(x) + Derivative(y(x), (x, 2)),0) 
ics = {y(1): 0} 
dsolve(ode,func=y(x),ics=ics)
 

$$y\left(x\right)=C_2\sin \left(x\right)+(-\frac{C_2\sin \left(1\right)}{\cos \left(1\right)}-\frac{x}{2}+\frac{1}{2})\cos \left(x\right)$$