4.9 HW 9

4.9.1 Problems listing

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4.9.2 Problem 1 (10.2.8)

Figure 4.25:Problem statement

Solution

4.9.2.1 Part 1

The ode to solve is $$\begin{array}{rl}\text{(1)}& x^{\prime \prime }(t)+2x^{\prime }(t)+x\left(t\right)& =0x(0)& =1\\ x^{\prime }(0)& =0\end{array}$$

This is a constant coefficient ODE. Assuming the solution has the form $x=A{e}^{\lambda t}$ and substituting this back in (1) gives the characteristic equation (the constant $A$ drops out)$$\begin{array}{rl}{\lambda }^2{e}^{\lambda t}+2\lambda e^{\lambda t}+e^{\lambda t}& =0\\ ({\lambda }^2+42\lambda +1)e^{\lambda t}& =0\end{array}$$

Since $e^{\lambda t}\ne 0$, the above gives$$\begin{array}{rl}{\lambda }^2+2\lambda +1& =0\\ {(\lambda +1)}^2& =0\end{array}$$

Therefore $\lambda =-1$. (double root). Since the root is double, then the basis solutions are $x_1(t)=e^{\lambda t},x_2\left(t\right)=t{e}^{\lambda t}$ and the general solution is a linear combination of these basis solutions. Therefore the general solution is$$\begin{array}{}\text{(2)}& x(t)=A{e}^{-t}+Bt{e}^{-t}\end{array}$$ The constants $A,B$ are found from initial conditions. At $t=0$ and using $x(0)=1$ gives$$\begin{array}{}\text{(3)}& 1=A\end{array}$$ Solution (2) becomes$$\begin{array}{}\text{(4)}& x(t)=e^{-t}+Bt{e}^{-t}\end{array}$$ Taking derivative of (4) gives$$x^{\prime }(t)=-e^{-t}+B{e}^t-Bt{e}^{-t}$$ Using $x^{\prime }(0)=0$ on the above gives$$\begin{array}{rl}0& =-1+B\\ \text{(5)}& B& =1\end{array}$$

Substituting (3,5) in (4) gives the final solution$$\begin{array}{rl}x(t)& =e^{-t}+t{e}^{-t}\\ & =(1+t)e^{-t}\end{array}$$

4.9.2.2 Part 2

The ode to solve is $$x^{\prime \prime \prime \prime }(t)+x(t)=0$$ As was done in the above part, substituting $x=A{e}^{\lambda t}$ in the above and simplifying gives the characteristic equation$${\lambda }^4+1=0$$ Hence the roots are ${\lambda }^4=-1$ or ${\lambda }^4=e^{-i\frac{\pi }{2}}$. There are 4 roots that divide the unit circle equally, each is $90$ degrees phase shifted (anti clockwise)from the other, starting from first root at phase $-\frac{\pi }{2}=-45$ degrees. Hence the roots are$$\begin{array}{rl}{\lambda }_1& =\cos (-45)+i\sin (-45)\\ {\lambda }_2& =\cos \left(45\right)+i\sin \left(45\right)\\ {\lambda }_3& =\cos \left(135\right)+i\sin \left(135\right)\\ {\lambda }_4& =\cos \left(225\right)+i\sin \left(225\right)\end{array}$$

or$$\begin{array}{rl}{\lambda }_1& =\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\\ {\lambda }_2& =\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\\ {\lambda }_3& =-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\\ {\lambda }_4& =-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\end{array}$$

Therefore the basis solutions are $$\begin{array}{rl}{x}_1(t)& =e^{(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})t}\\ x_2(t)& =e^{(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})t}\\ x_3(t)& =e^{(-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})t}\\ x_4(t)& =e^{(-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})t}\end{array}$$

The general solution is linear combination of the above basis solutions, which becomes$$\begin{array}{rl}x(t)& =c_1{e}^{(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})t}+c_2{e}^{(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})t}+c_3{e}^{(-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})t}+c_4{e}^{(-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})t}\\ & =c_1{e}^{\frac{\sqrt{2}}{2}t}{e}^{-i\frac{\sqrt{2}}{2}t}+c_2{e}^{\frac{\sqrt{2}}{2}t}{e}^{i\frac{\sqrt{2}}{2}t}+c_3{e}^{-\frac{\sqrt{2}}{2}t}{e}^{i\frac{\sqrt{2}}{2}t}+c_4{e}^{-\frac{\sqrt{2}}{2}t}{e}^{-i\frac{\sqrt{2}}{2}t}\\ & =e^{\frac{\sqrt{2}}{2}t}(c_1{e}^{-i\frac{\sqrt{2}}{2}t}+c_2{e}^{i\frac{\sqrt{2}}{2}t})+e^{-\frac{\sqrt{2}}{2}t}(c_3{e}^{i\frac{\sqrt{2}}{2}t}+c_4{e}^{-i\frac{\sqrt{2}}{2}t})\end{array}$$

Using Euler relation, the above can be rewritten as$$x(t)=e^{\frac{\sqrt{2}}{2}t}(c_1\sin \left(\frac{\sqrt{2}}{2}t\right)+c_2\cos \left(\frac{\sqrt{2}}{2}t\right))+e^{-\frac{\sqrt{2}}{2}t}(c_3\sin \left(\frac{\sqrt{2}}{2}t\right)+c_4\cos \left(\frac{\sqrt{2}}{2}t\right))$$

4.9.2.3 Part 3

The ode to solve is $$x^{\prime \prime \prime }(t)-3x^{\prime \prime }(t)-9x^{\prime }(t)-5x(t)=0$$ As was done in the above part, substituting $x=A{e}^{\lambda t}$ in the above and simplifying gives the characteristic equation$${\lambda }^3-3{\lambda }^2-9\lambda -5=0$$ Since one root is $5$, then the above can be written as$$(\lambda -5)\left(\mathrm{\Delta }\right)=0$$ Where $$\mathrm{\Delta }=\frac{{\lambda }^3-3{\lambda }^2-9\lambda -5}{\lambda -5}$$ Using long division gives$$\mathrm{\Delta }={(\lambda +1)}^2$$ Therefore the roots of the characteristic equation are$$\begin{array}{rl}{\lambda }_1& =5\\ {\lambda }_2& =-1\\ {\lambda }_3& =-1\end{array}$$

roots ${\lambda }_2,{\lambda }_3$ are the same. $\lambda =-1$ is a double root. Therefore the basis solutions are$$\begin{array}{rl}{x}_1(t)& =e^{5t}\\ x_2(t)& =e^t\\ x_3(t)& =t{e}^t\end{array}$$

Where $t$ multiplies the last basis $x_3(t)$ due to the double root. The general solution is linear combination of the above basis solutions, which gives$$\begin{array}{rl}x(t)& =c_1{x}_1(t)+c_2{x}_2\left(t\right)+c_3{x}_3(t)\\ & =c_1{e}^{5t}+c_2{e}^t+c_{3}t{e}^t\end{array}$$

4.9.2.4 Part 4

The ode to solve is $${(D+1)}^2(D^4-256)x(t)=0$$ This has the characteristic equation equation ${(\lambda +1)}^2({\lambda }^4-256)=0$. The roots of $({\lambda }^4-256)$ are given by ${\lambda }^4=256$. Let ${\lambda }^2=\omega$. Therefore ${\omega }^2=256$ which gives $\omega =\pm 16$.

When $\omega =16$, then ${\lambda }^2=16$ which gives $\lambda =\pm 4$ and when $\omega =-16$, then ${\lambda }^2=-16$ which gives $\lambda =\pm 4i.$

The other part ${(\lambda +1)}^2=0$ gives $\lambda =-1$, double root. Therefore the roots of the characteristic equation are$$\begin{array}{rl}{\lambda }_1& =4\\ {\lambda }_2& =-4\\ {\lambda }_3& =4i\\ {\lambda }_4& =-4i\\ {\lambda }_5& =-1\\ {\lambda }_6& =-1\end{array}$$

Root $\lambda =-1$ is a double root. Therefore the basis solutions as$$\begin{array}{rl}{x}_1(t)& =e^{4t}\\ x_2(t)& =e^{-4t}\\ x_3(t)& =e^{4it}\\ x_4(t)& =e^{-4it}\\ x_5(t)& =e^{-t}\\ x_6(t)& =t{e}^{-t}\end{array}$$

Where $t$ was multiplied by $e^{-t}$ in $x_6(t)$ since the root is double. The solution is linear combination of the above basis solutions, which gives$$\begin{array}{rl}x(t)& =c_1{x}_1(t)+c_2{x}_2\left(t\right)+c_3{x}_3(t)+c_4{x}_4(t)+c_5{x}_5(t)+c_6{x}_6(t)\\ & =c_1{e}^{4t}+c_2{e}^{-4t}+c_3{e}^{4it}+c_4{e}^{-4it}+c_5{e}^{-t}+c_{6}t{e}^{-t}\\ & =e^{-t}(c_5+t{c}_6)+c_1{e}^{4t}+c_2{e}^{-4t}+c_3\sin \left(4t\right)+c_4\cos \left(4t\right)\end{array}$$

Where Euler relation was used in the last step above to rewrite $c_3{e}^{4it}+c_4{e}^{-4it}$.

4.9.3 Problem 2 (10.2.11)

Figure 4.26:Problem statement

Solution

4.9.3.1 Part 1

The ode to solve is $$\begin{array}{}\text{(1)}& y^{\prime \prime }-y^{\prime }-2y=e^{2x}\end{array}$$ This is second order constant coefficients inhomogeneous ODE. The general solution is $$\begin{array}{}\text{(2)}& y(x)=y_h(x)+y_p(x)\end{array}$$ Where $y_h(x)$ is the solution to $y^{\prime \prime }-y^{\prime }-2y=0$ and $y_p(x)$ is any particular solution to $y^{\prime \prime }-y^{\prime }-2y=e^{2x}$. The homogenous solution is found using the characteristic polynomial method as was done in the above problems.  Substituting $y=A{e}^{\lambda x}$ in $y^{\prime \prime }-y^{\prime }-2y=0$ and simplifying gives$$\begin{array}{rl}{\lambda }^2-\lambda -2& =0\\ (\lambda +1)(\lambda -2)& =0\end{array}$$

The roots are ${\lambda }_1=-1,{\lambda }_2=2$. Therefore the basis solutions are $$\begin{array}{rl}\text{(3)}& y_1(x)& =e^{-x}{y}_1(x)& =e^{2x}\end{array}$$

Hence $y_h(x)$ is linear combination of the above, which gives$$y_h(x)=c_1{e}^{-x}+c_2{e}^{2x}$$ The particular solution is now found. Assuming $y_p=A{e}^{2x}$. But $e^{2x}$ is a basis solution of the homogeneous ode. Therefore $y_p$ is multiplied by $x$ giving $$y_p=Ax{e}^{2x}$$ Substituting this back in (1) and solving for $A$ gives$$\begin{array}{rl}{y}_p^{\prime }& =A{e}^{2x}+2Ax{e}^{2x}\\ y_p^{\prime \prime }& =2A{e}^{2x}+2A{e}^{2x}+4Ax{e}^{2x}\\ & =4A{e}^{2x}+4Ax{e}^{2x}\end{array}$$

Eq (1) becomes$$\begin{array}{rl}(4A{e}^{2x}+4Ax{e}^{2x})-(A{e}^{2x}+2Ax{e}^{2x})-2\left(Ax{e}^{2x}\right)& =e^{2x}\\ 4A{e}^{2x}+4Ax{e}^{2x}-A{e}^{2x}-2Ax{e}^{2x}-2Ax{e}^{2x}& =e^{2x}\\ 4A+4Ax-A-2Ax-2Ax& =1\\ 3A& =1\\ A& =\frac{1}{3}\end{array}$$

Hence the particular solution is$$y_p(x)=\frac{1}{3}x{e}^{2x}$$ Therefore from (2) the general solution is$$\begin{array}{}\text{(4)}& y(x)=c_1{e}^{-x}+c_2{e}^{2x}+\frac{1}{3}x{e}^{2x}\end{array}$$ $c_1,c_2$ are now found from initial conditions. At $x=0$, (4) becomes$$\begin{array}{}\text{(5)}& 1=c_1+c_2\end{array}$$ Taking derivative of (4) gives$$y^{\prime }(x)=-c_1{e}^{-x}+2c_2{e}^{2x}+\frac{1}{3}{e}^{2x}+\frac{2}{3}x{e}^{2x}$$ At $x=0$ the above gives$$\begin{array}{}\text{(6)}& 0=-c_1+2c_2+\frac{1}{3}\end{array}$$ Eq (5,6) are now solved for $c_1,c_2$. From (5) $$c_1=1-c_2$$ Substituting this back in (6) gives$$\begin{array}{rl}0& =-(1-c_2)+2c_2+\frac{1}{3}\\ c_2& =\frac{2}{9}\end{array}$$

Therefore $c_1=1-\frac{2}{9}=\frac{7}{9}$. The final solution (4) becomes$$y(x)=\frac{7}{9}{e}^{-x}+\frac{2}{9}{e}^{2x}+\frac{1}{3}x{e}^{2x}$$

4.9.3.2 Part 2

The ode to solve is $$\begin{array}{}\text{(1)}& y^{\prime \prime }-2y^{\prime }+y=2\cos x\end{array}$$ This is second order constant coefficients inhomogeneous ODE. Hence the general solution is $$\begin{array}{}\text{(2)}& y(x)=y_h(x)+y_p(x)\end{array}$$ Where $y_h(x)$ is the solution to $y^{\prime \prime }-2y^{\prime }+y=0$ and $y_p(x)$ is any particular solution to $y^{\prime \prime }-2y^{\prime }+y=2\cos x$. The homogenous is found using the characteristic polynomial method.  Substituting $y=A{e}^{\lambda x}$ in $y^{\prime \prime }-2y^{\prime }+y=0$ and simplifying gives$$\begin{array}{rl}{\lambda }^2-2\lambda +1& =0\\ (\lambda -1)(\lambda -1)& =0\end{array}$$

roots are ${\lambda }_1=1,{\lambda }_2=1$. (double root). The basis solutions are therefore$$\begin{array}{rl}\text{(3)}& y_1(x)& =e^x{y}_1(x)& =x{e}^x\end{array}$$

$y_h(x)$ is linear combination of the the above which gives$$y_h(x)=c_1{e}^x+c_{2}x{e}^x$$ The particular solution is now found. Assuming $y_p=A\cos x$. Taking all derivatives of this solution gives the set $\{\cos x,\sin x\}$. Therefore $$y_p=A\cos x+B\sin x$$ Substituting this back in (1) to solve for $A,B$ gives$$\begin{array}{rl}{y}_p^{\prime }& =-A\sin x+B\cos x\\ y_p^{\prime \prime }& =-A\cos x-B\sin x\end{array}$$

Hence (1) becomes$$\begin{array}{rl}{y}_p^{\prime \prime }-2y_p^{\prime }+y_p& =2\cos x\\ (-A\cos x-B\sin x)-2(-A\sin x+B\cos x)+(A\cos x+B\sin x)& =2\cos x\\ -A\cos x-B\sin x+2A\sin x-2B\cos x+A\cos x+B\sin x& =2\cos x\\ \cos x(-A-2B+A)+\sin x(-B+2A+B)& =2\cos x\\ -2B\cos x+2A\sin x& =2\cos x\end{array}$$

Hence $A=0$ and $B=-1$. Therefore the particular solution is$$y_p(x)=-\sin x$$ Eq (2) becomes$$\begin{array}{}\text{(4)}& y(x)=c_1{e}^x+c_{2}x{e}^x-\sin x\end{array}$$ $c_1,c_2$ are now found from initial conditions. At $x=0$, (4) becomes$$\begin{array}{}\text{(5)}& 1=c_1\end{array}$$ The solution (4) becomes$$\begin{array}{}\text{(6)}& y(x)=e^x+c_{2}x{e}^x-\sin x\end{array}$$ Taking derivative of (6) gives$$y^{\prime }(x)=e^x+c_2{e}^x+c_{2}x{e}^x-\cos x$$ At $x=0$ the above gives$$\begin{array}{}\text{(6)}& 0=1+c_2-1\end{array}$$ Therefore $c_2=0$ and now Eq (6) gives the final solution as$$y(x)=e^x-\sin x$$

4.9.3.3 Part 3

The ode to solve is $$\begin{array}{}\text{(1)}& y^{\prime \prime }+16y=16\cos 4x\end{array}$$ This is second order constant coefficients inhomogeneous ODE. Hence the general solution is $$\begin{array}{}\text{(2)}& y(x)=y_h(x)+y_p(x)\end{array}$$ Where $y_h(x)$ is the solution to $y^{\prime \prime }+16y=0$ and $y_p(x)$ is any particular solution to $y^{\prime \prime }+16y=16\cos 4x$. The homogenous is found using the characteristic polynomial method.  Substituting $y=A{e}^{\lambda x}$ in $y^{\prime \prime }+16y=0$ and simplifying gives$$\begin{array}{rl}{\lambda }^2+16& =0\\ \lambda & =\pm 4i\end{array}$$

The roots are ${\lambda }_1=4i,{\lambda }_2=-4i$. The basis solutions are therefore$$\begin{array}{rl}\text{(3)}& y_1(x)& =e^{i4x}{y}_2(x)& =e^{-i4x}\end{array}$$

Therefore $y_h(x)$ is linear combination of the the above. $$y_h(x)=c_1{e}^{i4x}+c_2{e}^{-i4x}$$ Which can be written, using Euler formula as$$y_h(x)=c_1\cos 4x+c_2\sin 4x$$ The particular solution is now found. Assuming $y_p=A\cos 4x$. Taking all derivatives of this, the basis for $y_p$ becomes $\{\cos 4x,\sin 4x\}$. But $\cos 4x$ is a basis of $y_h$. Therefore this set is multiplied by $x$. The whole set is multiplied by $x$ and not just $\cos 4x$ because the set was generated by taking derivative of $\cos 4x$.

The basis set for $y_p$ now becomes $\{x\cos 4x,x\sin 4x\}$. Hence $y_p$ is linear combination of these basis, giving trial $y_p$ as$$\begin{array}{}\text{(4)}& y_p=Ax\cos 4x+Bx\sin 4x\end{array}$$ Therefore$$\begin{array}{rl}{y}_p^{\prime }& =(A\cos 4x-4Ax\sin 4x)+(B\sin 4x+4Bx\cos 4x)\\ y_p^{\prime \prime }& =(-4A\sin 4x-4A\sin 4x-16Ax\cos 4x)+(4B\cos 4x+4B\cos 4x-16Bx\sin 4x)\\ & =-8A\sin 4x-16Ax\cos 4x+8B\cos 4x-16Bx\sin 4x\end{array}$$

Substituting the above back in (1) gives$$\begin{array}{rl}(-8A\sin 4x-16Ax\cos 4x+8B\cos 4x-16Bx\sin 4x)+16(Ax\cos 4x+Bx\sin 4x)& =16\cos 4x\\ \sin 4x(-8A-16Bx+16Bx)+\cos 4x(-16Ax+8B+16Ax)& =16\cos 4x\end{array}$$

Hence$$\begin{array}{rl}-16Ax+8B+16Ax& =16\\ -8A-16Bx+16Bx& =0\end{array}$$

Or$$\begin{array}{rl}8B& =16\\ -8A& =0\end{array}$$

First equation gives $B=2$. Second equation gives $A=0$. Therefore the particular solution (4) becomes$$y_p=2x\sin 4x$$ From (2), the general solution becomes$$\begin{array}{rl}y(x)& =y_h(x)+y_p(x)\\ \text{(5)}& & =c_1\cos 4x+c_2\sin 4x+2x\sin 4x\end{array}$$

$c_1,c_2$ are now found from initial conditions. At $x=0$, (5) becomes$$1=c_1$$ Hence the solution (5) becomes$$\begin{array}{}\text{(6)}& y(x)=\cos 4x+c_2\sin 4x+2x\sin 4x\end{array}$$ Taking derivative of the above$$y^{\prime }(x)=-4\sin 4x+4c_2\cos 4x+2\sin 4x+8x\cos 4x$$ At $t=0$ the above gives$$0=4c_2$$ Hence $c_2=0$ and the final solution (6) becomes$$y(x)=\cos 4x+2x\sin 4x$$

4.9.3.4 Part 4

The ode to solve is $$\begin{array}{}\text{(1)}& y^{\prime \prime }-y=\cosh x\end{array}$$ This is second order constant coefficients inhomogeneous ODE. Hence the general solution is $$\begin{array}{}\text{(2)}& y(x)=y_h(x)+y_p(x)\end{array}$$ Where $y_h(x)$ is the solution to $y^{\prime \prime }+y=0$ and $y_p(x)$ is any particular solution to $y^{\prime \prime }+y=\cosh x$. The homogenous is found using the characteristic polynomial method.  Substituting $y=A{e}^{\lambda x}$ in $y^{\prime \prime }+y=0$ and simplifying gives$$\begin{array}{rl}{\lambda }^2-1& =0\\ \lambda & =\pm 1\end{array}$$

roots are ${\lambda }_1=1,{\lambda }_2=-1$. The basis solutions are therefore$$\begin{array}{rl}\text{(3)}& y_1(x)& =e^x{y}_2(x)& =e^{-x}\end{array}$$

Therefore $y_h(x)$ is linear combination of the the above. $$y_h(x)=c_1{e}^x+c_2{e}^{-x}$$ Which can be written, using Euler formula as$$y_h(x)=c_1\cosh x+c_2\sinh x$$ The particular solution is now found. Assuming $y_p=A\cosh x$. Taking all derivatives of this, the basis for $y_p$ becomes $\{\cosh x,\sinh x\}$. But $\cosh x$ is basis of $y_h$. Therefore this set is multiplied by $x$. The whole set is multiplied by $x$ and not just $\cosh x$ because the set was generated by taking derivative of $\cosh x$.

The basis set for $y_p$ becomes $\{x\cosh x,x\sinh x\}$. Hence $y_p$ is linear combination of these basis, giving trial $y_p$ as$$\begin{array}{}\text{(4)}& y_p=Ax\cosh x+Bx\sinh x\end{array}$$ Therefore$$\begin{array}{rl}{y}_p^{\prime }& =A\cosh x+Ax\sinh x+B\sinh x+Bx\cosh x\\ y_p^{\prime \prime }& =A\sinh x+A\sinh x+Ax\cosh x+B\cosh x+B\cosh x+Bx\sinh x\\ & =2A\sinh x+Ax\cosh x+2B\cosh x+Bx\sinh x\end{array}$$

Substituting the above back in (1) gives$$\begin{array}{rl}(2A\sinh x+Ax\cosh x+2B\cosh x+Bx\sinh x)-(Ax\cosh x+Bx\sinh x)& =\cosh x\\ \sinh x(2A+Bx-Bx)+\cosh x(Ax+2B-Ax)& =\cosh x\end{array}$$

Hence$$\begin{array}{rl}2B& =1\\ 2A& =0\end{array}$$

Therefore $B=\frac{1}{2},A=0$ and (4) becomes$$y_p=\frac{1}{2}x\sinh x$$ From (2), the general solution becomes$$\begin{array}{rl}y(x)& =y_h(x)+y_p(x)\\ \text{(5)}& & =c_1\cosh x+c_2\sinh x+\frac{1}{2}x\sinh x\end{array}$$

$c_1,c_2$ are now found from initial conditions. At $x=0$, (5) becomes$$1=c_1$$ Hence the solution (5) becomes$$\begin{array}{}\text{(6)}& y(x)=\cosh x+c_2\sinh x+\frac{1}{2}x\sinh x\end{array}$$ Taking derivative of the above$$y^{\prime }(x)=\sinh x+c_2\cosh x+\frac{1}{2}\sinh x+\frac{1}{2}x\cosh x$$ At $t=0$ the above gives$$0=c_2\cosh x$$ Hence $c_2=0$ and the final solution (6) becomes$$y(x)=\cosh x+\frac{1}{2}x\sinh x$$

4.9.4 Problem 3 (10.3.5)

Solve $x^2{y}^{\prime }+2xy=\sinh x$ with $y(1)=2$

Solution

Dividing by $x\ne 0$$$y^{\prime }+2\frac{y}{x}=\frac{\sinh x}{x^2}$$ The integrating factor is $I=e^{\int \frac{2}{x}dx}=e^{2\ln x}=x^2$. Multiplying both sides by this integration factor makes the left side a complete differential$$\begin{array}{rl}\frac{d}{dx}\left(y{x}^2\right)& =x^2\frac{\sinh x}{x^2}\\ \frac{d}{dx}\left(y{x}^2\right)& =\sinh x\end{array}$$

Integrating gives$$\begin{array}{rl}y{x}^2& =\int \sinh xdx+C\\ y{x}^2& =\cosh x+C\\ \text{(1)}& y& =\frac{\cosh x}{x^2}+\frac{C}{x^2}\end{array}$$

At $x=1$ the above becomes$$\begin{array}{rl}2& =\cosh 1+C\\ C& =2-\cosh 1\end{array}$$

Hence the solution (1) becomes$$\begin{array}{rl}y(x)& =\frac{\cosh x}{x^2}+\frac{2-\cosh 1}{x^2}\\ & =\frac{1}{x^2}(\cosh x+2-\cosh 1)\end{array}$$

Where $x\ne 0$

4.9.5 Problem 4 (10.3.8)

Solve $$(1+x^2)y^{\prime }=1+xy$$ Solution

$$\begin{array}{rl}{y}^{\prime }& =\frac{1+xy}{1+x^2}\\ & =\frac{1}{1+x^2}+\frac{xy}{1+x^2}\end{array}$$

Therefore$$\begin{array}{}\text{(1)}& y^{\prime }-y\frac{x}{1+x^2}=\frac{1}{1+x^2}\end{array}$$ This is linear in $y$ first order ODE. It has the form $y^{\prime }+p\left(x\right)y=q(x)$. The integration factor is $$\begin{array}{rl}I& =e^{\int p(x)dx}\\ & =e^{-\int \frac{x}{1+x^2}dx}\end{array}$$

But $\int \frac{x}{1+x^2}dx=\frac{1}{2}\ln (1+x^2)$. Therefore$$\begin{array}{rl}I& =e^{-\frac{1}{2}\ln (1+x^2)}\\ & =e^{\ln {(1+x^2)}^{-\frac{1}{2}}}\\ & ={(1+x^2)}^{-\frac{1}{2}}\\ & =\frac{1}{\sqrt{1+x^2}}\end{array}$$

Multiplying both sides of (1) by this integrating factor makes the left side a complete differential$$\begin{array}{rl}\frac{d}{dx}\left(y\frac{1}{\sqrt{1+x^2}}\right)& =\frac{1}{\sqrt{1+x^2}}\frac{1}{1+x^2}\\ \frac{d}{dx}\left(y\frac{1}{\sqrt{1+x^2}}\right)& =\frac{1}{{(1+x^2)}^{\frac{3}{2}}}\\ & ={(1+x^2)}^{-\frac{3}{2}}\end{array}$$

Integrating gives$$\begin{array}{}\text{(2)}& y\frac{1}{\sqrt{1+x^2}}=\int {(1+x^2)}^{-\frac{3}{2}}dx+C\end{array}$$ To integrate $\int \frac{1}{{(1+x^2)}^{\frac{3}{2}}}dx$, let $x=\tan u$, then $dx=(1+{\tan }^{2}u)du$. Hence$$\begin{array}{rl}\int \frac{1}{{(1+x^2)}^{\frac{3}{2}}}dx& =\int \frac{1}{{(1+{\tan }^{2}u)}^{\frac{3}{2}}}(1+{\tan }^{2}u)du\\ & =\int \frac{1}{{(1+{\tan }^{2}u)}^{\frac{1}{2}}}du\\ & =\int \frac{1}{{(1+\frac{{\sin }^{2}u}{{\cos }^{2}u})}^{\frac{1}{2}}}du\\ & =\int \frac{\cos u}{{({\cos }^{2}u+{\sin }^{2}u)}^{\frac{1}{2}}}du\\ & =\int \cos udu\\ & =\sin u\end{array}$$

But $\sin u=\frac{\frac{\sin u}{\cos u}}{\sqrt{1+\frac{{\sin }^{2}u}{{\cos }^{2}u}}}=\frac{\tan u}{\sqrt{1+{\tan }^{2}u}}=\frac{x}{\sqrt{1+x^2}}$. Hence $$\int \frac{1}{{(1+x^2)}^{\frac{3}{2}}}dx=\frac{x}{\sqrt{1+x^2}}$$ Therefore the final solution (2) becomes$$\begin{array}{rl}y\frac{1}{\sqrt{1+x^2}}& =\frac{x}{\sqrt{1+x^2}}+C\\ \text{(3)}& y& =x+C\sqrt{1+x^2}\end{array}$$

4.9.6 Problem 5 (10.3.9)

Solve (a) $y^{\prime }+xy=x{y}^2$ (b) $3x{y}^{\prime }+y+x^2{y}^4=0$

Solution

4.9.6.1 Part a

The ode has the form$$y^{\prime }+p(x)y=q(x)y^m$$ Where $p(x)=x,q(x)=x$ and $m=2$. Therefore this is Bernoulli ODE. The first step is to divide throughout by $y^m=y^2$ which gives$$\begin{array}{}\text{(1)}& \frac{y^{\prime }}{y^2}+p(x)y^{-1}=q(x)\end{array}$$ Setting $$\begin{array}{}\text{(2)}& v(x)=y^{-1}\end{array}$$ Taking derivatives of the above w.r.t. $x$ gives $$\begin{array}{}\text{(3)}& v^{\prime }(x)=\frac{-1}{y^2}{y}^{\prime }(x)\end{array}$$ Substituting (2,3) into (1) gives$$-v^{\prime }(x)+p(x)v(x)=q\left(x\right)$$ But here $p(x)=x$ and $q(x)=x$. The above becomes$$\begin{array}{rl}-v^{\prime }(x)+xv(x)& =x\\ v^{\prime }(x)-xv(x)& =-x\end{array}$$

This is linear ODE in $v(x)$. The integrating factor is $e^{\int -xdx}=e^{-\frac{x^2}{2}}$. Multiplying both sides of the above by this integrating factor makes the left side a complete differential$$\frac{d}{dx}\left(v{e}^{-\frac{x^2}{2}}\right)=-x{e}^{-\frac{x^2}{2}}$$ Integrating gives$$\begin{array}{}\text{(4)}& v{e}^{-\frac{x^2}{2}}=-\int x{e}^{-\frac{x^2}{2}}dx+C\end{array}$$ To integrate $\int x{e}^{-\frac{x^2}{2}}dx$, let $u=x^2$. Then $du=2xdx$. Substituting gives$$\begin{array}{rl}\int x{e}^{-\frac{x^2}{2}}dx& =\int x{e}^{-\frac{u}{2}}\frac{du}{2x}\\ & =\frac{1}{2}\int e^{-\frac{u}{2}}du\\ & =\frac{1}{2}\frac{e^{-\frac{u}{2}}}{-\frac{1}{2}}\\ & =-e^{-\frac{u}{2}}\end{array}$$

But $u=x^2$. Therefore$$\int x{e}^{-\frac{x^2}{2}}dx=-e^{-\frac{x^2}{2}}$$ Substituting the above in (4) gives$$\begin{array}{rl}v{e}^{-\frac{x^2}{2}}& =e^{-\frac{x^2}{2}}+C\\ v& =1+e^{\frac{x^2}{2}}C\end{array}$$

But $v=y^{-1}$, therefore$$\begin{array}{rl}{y}^{-1}& =1+e^{\frac{x^2}{2}}C\\ y(x)& =\frac{1}{1+e^{\frac{x^2}{2}}C}\end{array}$$

Where $C$ is constant of integration.

4.9.6.2 Part b

The ode is$$3x{y}^{\prime }+y+x^2{y}^4=0$$ Dividing by $3x$ for $x\ne 0$ gives$$\begin{array}{rl}{y}^{\prime }+\frac{y}{3x}+\frac{x}{3}{y}^4& =0\\ y^{\prime }+\frac{1}{3x}y& =-\frac{x}{3}{y}^4\end{array}$$

Now this ODE has the Bernoulli form, $$y^{\prime }+p(x)y=q(x)y^m$$ Where $p(x)=\frac{1}{3x},q(x)=-\frac{x}{3}$ and $m=4$. Therefore this is Bernoulli ODE. The first step is to divide throughout by $y^m=y^4$ which gives$$\begin{array}{}\text{(1)}& \frac{y^{\prime }}{y^4}+p(x)y^{-3}=q(x)\end{array}$$ Setting $$\begin{array}{}\text{(2)}& v(x)=y^{-3}\end{array}$$ Taking derivatives of the above w.r.t. $x$ gives $$\begin{array}{}\text{(3)}& v^{\prime }(x)=\frac{-3}{y^4}{y}^{\prime }(x)\end{array}$$ Substituting (2,3) into (1) gives$$-\frac{1}{3}{v}^{\prime }(x)+p(x)v(x)=q(x)$$ But here $p(x)=\frac{1}{3x},q(x)=-\frac{x}{3}$. The above becomes$$\begin{array}{rl}-\frac{1}{3}{v}^{\prime }(x)+\frac{1}{3x}v(x)& =-\frac{x}{3}\\ v^{\prime }(x)-\frac{1}{x}v(x)& =x\end{array}$$

This is linear in $v(x)$. The integrating factor is $e^{-\int \frac{1}{x}dx}=e^{-\ln x}=\frac{1}{x}$. Multiplying both sides of the above by this integrating factor make the left side a complete differential$$\frac{d}{dx}\left(v\frac{1}{x}\right)=1$$ Integrating gives$$\begin{array}{rl}v\frac{1}{x}& =x+C\\ \text{(4)}& v& =x^2+xC\end{array}$$

But $v(x)=y^{-3}$. Therefore the above becomes$$\begin{array}{rl}{y}^{-3}& =x^2+xC\\ y^3(x)& =\frac{1}{x^2+xC}\end{array}$$

Or$$y(x)={(x^2+xC)}^{-\frac{1}{3}}$$

4.9.7 key solution for HW 9

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