In Problems 5–8, determine whether or not each indicated set of functions is a subspace of the space \(F\) of all real-valued functions on \(\mathbb {R} \).
The set of all \(f\) such that \(f\left ( 0\right ) =0\) and \(f\left ( 1\right ) =1\)
Solution
Let \(f,g\) be two functions such that \(f\left ( 0\right ) =0,g\left ( 0\right ) =0\) and \(f\left ( 1\right ) =1,g\left ( 1\right ) =1\) in \(F\). Let us check if it is closed under addition\[ f\left ( 0\right ) +g\left ( 0\right ) =0+0=0 \] OK.\[ f\left ( 1\right ) +g\left ( 1\right ) =1+1=2\neq 1 \] Hence not closed under addition. Therefore not a subspace.
In Problems 9–12, a condition on the coefficients of a polynomial \(a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}\) is given. Determine whether or not the set of all such polynomials satisfying this condition is a subspace of the space \(P\) of all polynomials\[ a_{0}=a_{1}=0 \] Solution
Let \begin {align*} p_{1}\left ( x\right ) & =a_{2}x^{2}+a_{3}x^{3}\\ p_{2}\left ( x\right ) & =b_{2}x^{2}+b_{3}x^{3} \end {align*}
Checking if closed under scalar multiplication. Let \(c\) be some scalar. Hence\begin {align*} cp_{1}\left ( x\right ) & =c\left ( a_{2}x^{2}+a_{3}x^{3}\right ) \\ & =\left ( ca_{2}\right ) x^{2}+\left ( ca_{3}\right ) x^{3}\\ & =A_{2}x^{2}+A_{3}x^{3} \end {align*}
Therefore closed. Now checking if closed under addition.\begin {align*} p_{1}\left ( x\right ) +p_{2}\left ( x\right ) & =a_{2}x^{2}+a_{3}x^{3}+b_{2}x^{2}+b_{3}x^{3}\\ & =\left ( a_{2}+b_{2}\right ) x^{2}+\left ( a_{3}+b_{3}\right ) x^{3}\\ & =A_{2}x^{2}+A_{3}x^{3} \end {align*}
Therefore Closed under addition. Also the zero polynomial in included when \(a_{2}=a_{3}=0\).
Therefore this is a subspace.
In Problems 1 through 12, verify by substitution that each given function is a solution of the given differential equation. Throughout these problems, primes denote derivatives with respect to \(x\).\begin {align} y^{\prime } & =y+2e^{-x}\tag {A}\\ y & =e^{x}-e^{-x}\nonumber \end {align}
Solution
Using the solution given, we see that\begin {align} y^{\prime } & =e^{x}-\left ( -e^{-x}\right ) \nonumber \\ & =e^{x}+e^{-x}\tag {1} \end {align}
Substituting (1) into EQ. (A) gives\begin {align*} e^{x}+e^{-x} & =\left ( e^{x}-e^{-x}\right ) +2e^{-x}\\ e^{x}+e^{-x} & =e^{x}+e^{-x}\\ 0 & =0 \end {align*}
Hence the solution given satisfies the ODE.
In Problems 17 through 26, first verify that \(y\left ( x\right ) \) satisfies the given differential equation. Then determine a value of the constant \(C\) so that \(y\left ( x\right ) \) satisfies the given initial condition. Use a computer or graphing calculator (if desired) to sketch several typical solutions of the given differential equation, and highlight the one that satisfies the given initial condition.\begin {align} y^{\prime }+y & =0\tag {A}\\ y\left ( x\right ) & =Ce^{-x}\nonumber \\ y\left ( 0\right ) & =2\nonumber \end {align}
Solution
Using the solution given, we see that\begin {equation} y^{\prime }=-Ce^{-x}\tag {1} \end {equation} Substituting (1) into EQ. (A) gives\begin {align*} -Ce^{-x}+Ce^{-x} & =0\\ 0 & =0 \end {align*}
Hence the solution gives satisfies the ODE.
When \(x=0\) the solution becomes\begin {align*} 2 & =Ce^{-\left ( 0\right ) }\\ & =C \end {align*}
Hence \(C=2\) and the particular solution becomes\[ y\left ( x\right ) =2e^{-x}\] The following are some solutions plots for different \(C\)
restart; f:=(x,c)->c*exp(-x) p1:=plot(f(x,2),x=-5..5,gridlines=true,view=[-6..6, -6..6],color=red): p2:=plot(f(x,4),x=-5..5,gridlines=true,view=[-6..6, -6..6],color=blue): p3:=plot(f(x,-2),x=-5..5,gridlines=true,view=[-6..6, -6..6],color=green): p4:=plot(f(x,-4),x=-5..5,gridlines=true,view=[-6..6, -6..6],color=black): T:=plots:-textplot([[.5,2,"(0,2)"]], font=[times,16],tickmarks=NULL): plots:-display([p1,p2,p3,p4,T]);
A homogeneous second-order linear differential equation, two functions \(y_{1}\) and \(y_{2}\), and a pair of initial conditions are given. First verify that \(y_{1}\) and \(y_{2}\) are solutions of the differential equation. Then find a particular solution of the form \(y=c_{1}y_{1}+c_{2}y_{2}\) that satisfies the given initial conditions. Primes denote derivatives with respect to \(x\).\begin {align} y^{\prime \prime }+4y & =0\tag {1}\\ y_{1} & =\cos 2x\nonumber \\ y_{2} & =\sin 2x\nonumber \\ y\left ( 0\right ) & =3\nonumber \\ y^{\prime }\left ( 0\right ) & =8\nonumber \end {align}
Solution
Checking if \(y_{1}\left ( x\right ) \) is a solution. Since\begin {align} y_{1}^{\prime } & =-2\sin 2x\tag {2}\\ y_{1}^{\prime \prime } & =-4\cos 2x\tag {3} \end {align}
Substituting the above equations back into (1) gives\begin {align*} \left ( -4\cos 2x\right ) +4\cos 2x & =0\\ 0 & =0 \end {align*}
Hence \(y_{1}\) is a solution. We do the same for \(y_{2}\)\begin {align} y_{2}^{\prime } & =2\cos 2x\tag {4}\\ y_{2}^{\prime \prime } & =-4\sin 2x\tag {5} \end {align}
Substituting (4,5) back into (1) gives\begin {align*} \left ( -4\sin 2x\right ) +4\left ( \sin 2x\right ) & =0\\ 0 & =0 \end {align*}
Hence \(y_{2}\) is a solution. Let general solution be\begin {align} y\left ( x\right ) & =c_{1}y_{1}\left ( x\right ) +c_{2}y_{2}\left ( x\right ) \nonumber \\ & =c_{1}\cos 2x+c_{2}\sin 2x\tag {6} \end {align}
Applying the first initial conditions \(y\left ( 0\right ) =3\) in (6) gives\[ 3=c_{1}\] Hence (6) now becomes\begin {equation} y\left ( x\right ) =3\cos 2x+c_{2}\sin 2x\tag {7} \end {equation} Taking derivative of the above gives\[ y^{\prime }\left ( x\right ) =-6\sin 2x+2c_{2}\cos 2x \] Applying the second initial conditions \(y^{\prime }\left ( 0\right ) =8\) in the above gives\begin {align*} 8 & =2c_{2}\\ c_{2} & =4 \end {align*}
Therefore the general solution (6) becomes\begin {equation} \fbox {$y\left ( x\right ) =3\cos 2x+4\sin 2x$}\tag {8} \end {equation}
A homogeneous second-order linear differential equation, two functions \(y_{1}\) and \(y_{2}\), and a pair of initial conditions are given. First verify that \(y_{1}\) and \(y_{2}\) are solutions of the differential equation. Then find a particular solution of the form \(y=c_{1}y_{1}+c_{2}y_{2}\) that satisfies the given initial conditions. Primes denote derivatives with respect to \(x\).\begin {align} y^{\prime \prime }-3y^{\prime }+2y & =0\tag {1}\\ y_{1} & =e^{x}\nonumber \\ y_{2} & =e^{2x}\nonumber \\ y\left ( 0\right ) & =1\nonumber \\ y^{\prime }\left ( 0\right ) & =0\nonumber \end {align}
Solution
Checking if \(y_{1}\left ( x\right ) \) is a solution. Since\begin {align} y_{1}^{\prime } & =e^{x}\tag {2}\\ y_{1}^{\prime \prime } & =e^{x}\tag {3} \end {align}
Substituting the above equations back into (1) gives\begin {align*} e^{x}-3e^{x}+2e^{x} & =0\\ 0 & =0 \end {align*}
Hence \(y_{1}\) is a solution. We do the same for \(y_{2}\)\begin {align} y_{2}^{\prime } & =2e^{2x}\tag {4}\\ y_{2}^{\prime \prime } & =4e^{2x}\tag {5} \end {align}
Substituting (4,5) back into (1) gives\begin {align*} \left ( 4e^{2x}\right ) -3\left ( 2e^{2x}\right ) +2\left ( e^{2x}\right ) & =0\\ 4e^{2x}-6e^{2x}+2e^{2x} & =0\\ 0 & =0 \end {align*}
Hence \(y_{2}\) is a solution. Let general solution be\begin {align} y\left ( x\right ) & =c_{1}y_{1}+c_{2}y_{2}\nonumber \\ & =c_{1}e^{x}+c_{2}e^{2x}\tag {6} \end {align}
Applying the first initial conditions \(y\left ( 0\right ) =1\) in (6) gives\begin {equation} 1=c_{1}+c_{2}\tag {7} \end {equation} Taking derivative of Eq. (6) gives\[ y^{\prime }\left ( x\right ) =c_{1}e^{x}+2c_{2}e^{2x}\] Applying the second initial conditions \(y^{\prime }\left ( 0\right ) =0\) in the above gives\begin {equation} 0=c_{1}+2c_{2}\tag {8} \end {equation} We have two equations (7,8) to solve for the 2 unknowns \(c_{1},c_{2}\). (7)-(8) gives\[ c_{2}=-1 \] Hence from (7) \(c_{1}=1-c_{2}=1+1=2\). Therefore the solution (6) now becomes\[ y\left ( x\right ) =2e^{x}-e^{2x}\]
A homogeneous second-order linear differential equation, two functions \(y_{1}\) and \(y_{2}\), and a pair of initial conditions are given. First verify that \(y_{1}\) and \(y_{2}\) are solutions of the differential equation. Then find a particular solution of the form \(y=c_{1}y_{1}+c_{2}y_{2}\) that satisfies the given initial conditions. Primes denote derivatives with respect to \(x\).\begin {align} y^{\prime \prime }+y^{\prime } & =0\tag {1}\\ y_{1} & =1\nonumber \\ y_{2} & =e^{-x}\nonumber \\ y\left ( 0\right ) & =-2\nonumber \\ y^{\prime }\left ( 0\right ) & =8\nonumber \end {align}
Solution
Checking if \(y_{1}\left ( x\right ) \) is a solution. Since\begin {align} y_{1}^{\prime } & =0\tag {2}\\ y_{1}^{\prime \prime } & =0\tag {3} \end {align}
Substituting the above equations back into (1) gives\begin {align*} 0+0 & =0\\ 0 & =0 \end {align*}
Hence \(y_{1}\) is a solution. We do the same for \(y_{2}\)\begin {align} y_{2}^{\prime } & =-e^{-x}\tag {4}\\ y_{2}^{\prime \prime } & =e^{-x}\tag {5} \end {align}
Substituting (4,5) back into (1) gives\begin {align*} \left ( e^{-x}\right ) -e^{-x} & =0\\ 0 & =0 \end {align*}
Hence \(y_{2}\) is a solution. Let general solution be\begin {align} y\left ( x\right ) & =c_{1}y_{1}+c_{2}y_{2}\nonumber \\ & =c_{1}+c_{2}e^{-x}\tag {6} \end {align}
Applying the first initial conditions \(y\left ( 0\right ) =-2\) in (6) gives\begin {equation} -2=c_{1}+c_{2}\tag {7} \end {equation} Taking derivative of Eq. (6) gives\[ y^{\prime }\left ( x\right ) =-c_{2}e^{-x}\] Applying the second initial conditions \(y^{\prime }\left ( 0\right ) =8\) in the above gives\begin {align} 8 & =-c_{2}\nonumber \\ c_{2} & =-8\tag {8} \end {align}
Hence from (7)\begin {align*} -2 & =c_{1}+c_{2}\\ & =c_{1}-8\\ c_{1} & =6 \end {align*}
Therefore the solution (6) now becomes\begin {align*} y\left ( x\right ) & =c_{1}+c_{2}e^{-x}\\ & =6-8e^{-x} \end {align*}
Apply Theorems 5 and 6 to find general solutions of the differential equations given in Problems 33 through 42. Primes denote derivatives with respect to \(x\).\[ y^{\prime \prime }-3y^{\prime }+2y=0 \] Solution
The characteristic equation is \begin {align*} r^{2}-3r+2 & =0\\ \left ( r-1\right ) \left ( r-2\right ) & =0 \end {align*}
Hence the roots are \(r_{1}=1,r_{2}=2\). Therefore the general solution is\begin {align*} y\left ( x\right ) & =Ae^{r_{1}x}+Be^{r_{2}x}\\ & =Ae^{x}+Be^{2x} \end {align*}
Where \(A,B\) are the constants of integrations which are found from initial conditions.
Apply Theorems 5 and 6 to find general solutions of the differential equations given in Problems 33 through 42. Primes denote derivatives with respect to \(x\).\[ y^{\prime \prime }+5y^{\prime }=0 \] Solution
The characteristic equation is \begin {align*} r^{2}+5r & =0\\ \left ( r+5\right ) r & =0 \end {align*}
Hence the roots are \(r_{1}=0,r_{2}=-5\). Therefore the general solution is\begin {align*} y\left ( x\right ) & =Ae^{r_{1}x}+Be^{r_{2}x}\\ & =A+Be^{-5x} \end {align*}
Where \(A,B\) are the constants of integrations which are found from initial conditions.
Apply Theorems 5 and 6 to find general solutions of the differential equations given in Problems 33 through 42. Primes denote derivatives with respect to \(x\).\[ 4y^{\prime \prime }+4y^{\prime }+y=0 \] Solution
The characteristic equation is \begin {align*} 4r^{2}+4r+1 & =0\\ r^{2}+r+\frac {1}{4} & =0\\ \left ( r+\frac {1}{2}\right ) ^{2} & =0 \end {align*}
Hence the root is \(r=-\frac {1}{2}\). A double root. Therefore the general solution is\begin {align*} y\left ( x\right ) & =Ae^{rx}+Bxe^{rx}\\ & =Ae^{-\frac {1}{2}x}+Bxe^{-\frac {1}{2}x} \end {align*}
Where \(A,B\) are the constants of integrations which are found from initial conditions.
Let \(P_{2}\) be subspace of polynomials of degree at most \(2\). So elements of \(P_{2}\) look like \(a_{0}+a_{1}x+a_{2}x^{2}\). Show that \(\left \{ 3+x,1+x+x^{2},x-2x^{2}\right \} \) is basis for \(P_{2}\)
Solution
Assuming these are basis, then we can write\[ a_{0}+a_{1}x+a_{2}x^{2}=c_{1}\left ( 3+x\right ) +c_{2}\left ( 1+x+x^{2}\right ) +c_{3}\left ( x-2x^{2}\right ) \] For constants \(c_{1},c_{2},c_{3}\). If we can find unique solution for the \(c_{i}\) then these are basis. The above becomes\begin {align*} a_{0}+a_{1}x+a_{2}x^{2} & =3c_{1}+c_{2}+xc_{1}+xc_{2}+xc_{3}+x^{2}c_{2}-2x^{2}c_{3}\\ & =\left ( 3c_{1}+c_{2}\right ) +x\left ( c_{1}+c_{2}+c_{3}\right ) +x^{2}\left ( c_{2}-2c_{3}\right ) \end {align*}
Comparing coefficients gives the equations\begin {align*} a_{0} & =3c_{1}+c_{2}\\ a_{1} & =c_{1}+c_{2}+c_{3}\\ a_{2} & =c_{2}-2c_{3} \end {align*}
In Matrix form the above becomes\[\begin {bmatrix} 3 & 1 & 0\\ 1 & 1 & 1\\ 0 & 1 & -2 \end {bmatrix}\begin {bmatrix} c_{1}\\ c_{2}\\ c_{3}\end {bmatrix} =\begin {bmatrix} a_{0}\\ a_{1}\\ a_{2}\end {bmatrix} \] Augmented matrix is\[\begin {bmatrix} 3 & 1 & 0 & a_{0}\\ 1 & 1 & 1 & a_{1}\\ 0 & 1 & -2 & a_{2}\end {bmatrix} \] Replacing row 2 with row 1 gives\[\begin {bmatrix} 1 & 1 & 1 & a_{1}\\ 3 & 1 & 0 & a_{0}\\ 0 & 1 & -2 & a_{2}\end {bmatrix} \] \(R_{2}\rightarrow -3R_{1}+R_{2}\) gives\[\begin {bmatrix} 1 & 1 & 1 & a_{1}\\ 0 & -2 & -3 & a_{0}-3a_{1}\\ 0 & 1 & -2 & a_{2}\end {bmatrix} \] \(R_{3}\rightarrow R_{2}+2R_{3}\) gives\[\begin {bmatrix} 1 & 1 & 1 & a_{1}\\ 0 & -2 & -3 & a_{0}-3a_{1}\\ 0 & 0 & -7 & a_{0}-3a_{1}+2a_{2}\end {bmatrix} \] The matrix is now in Echelon form. We see that there are no free variables. Only leading variables \(c_{1},c_{2},c_{3}\). This implies we have unique solution. Which means we can solve for \(c_{1},c_{2},c_{3}\) in terms of \(a_{1},a_{2},a_{3}\). We are not asked to complete the solution, only to say if these are basis. So we can stop here.
This shows that \(\left \{ 3+x,1+x+x^{2},x-2x^{2}\right \} \) are basis for \(P_{2}.\)
Find the general solution for \(y^{\prime \prime }-25y=0\). What is the particular solution for \(y\left ( 0\right ) =a,y^{\prime }\left ( 0\right ) =b\)?
Solution
The characteristic equation is \begin {align*} r^{2}-25 & =0\\ r & =\pm 5 \end {align*}
Two distinct real roots \(r_{1}=5,r_{2}=-5\). Therefore the general solution is \begin {align} y\left ( x\right ) & =c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}\nonumber \\ & =c_{1}e^{5x}+c_{2}e^{-5x}\tag {1} \end {align}
Now we apply the initial conditions. The first one \(y\left ( 0\right ) =a\) applied to the above gives\begin {equation} a=c_{1}+c_{2}\tag {2} \end {equation} Taking derivative of (1) gives\[ y^{\prime }=5c_{1}e^{5x}-5c_{2}e^{-5x}\] Applying second initial conditions \(y^{\prime }\left ( 0\right ) =b\) to the above gives\begin {equation} b=5c_{1}-5c_{2}\tag {3} \end {equation} Multiplying (2) by \(5\) and adding the result to Eq (3) gives\begin {align*} 5a+b & =\left ( 5c_{1}+5c_{2}\right ) +\left ( 5c_{1}-5c_{2}\right ) \\ 5a+b & =10c_{1} \end {align*}
Hence\[ c_{1}=\frac {5a+b}{10}\] From (2) we now solve for \(c_{2}\)\begin {align*} a & =\frac {5a+b}{10}+c_{2}\\ c_{2} & =a-\frac {5a+b}{10}\\ & =\frac {a}{2}-\frac {b}{10} \end {align*}
Now that we found both constants, the particular solution becomes\begin {align*} y\left ( x\right ) & =c_{1}e^{5x}+c_{2}e^{-5x}\\ & =\left ( \frac {a}{2}+\frac {b}{10}\right ) e^{5x}+\left ( \frac {a}{2}-\frac {b}{10}\right ) e^{-5x} \end {align*}