Question

Could you help me find the general solution for these equations, please? Thank you

y'''+ y'' = y + x^5

y(y'') = y' + 8x^3

Answer 1

Equation #1

$$y'''+y''=y+x^5$$

1 STEP: Solve a homogeneous equation

$$y'''+y''-y=0\\$$

We are looking for a solution in the form

$$y(x)=e^{kx}\longrightarrow\,y''=k^2\cdot e^{kx}\,\,\,and\,\,\,y'''=k^3\cdot e^{kx}$$

Then,

$$k^3\cdot e^{kx}+k^2\cdot e^{kx}-e^{kx}=0\longrightarrow\\ k^3+k^2-1=0$$

the cubic equation has three solutions: one real and two complex

$$k_1=\frac{1}{3}\left(-1+\sqrt[3]{\frac{25}{2}-\frac{3\sqrt{69}}{2}}+\sqrt[3]{\frac{25+3\sqrt{69}}{2}}\right)\\[0.5cm] k_2=-\frac{1}{3}-\frac{1-i\sqrt{3}}{6}\left(\sqrt[3]{\frac{25}{2}-\frac{3\sqrt{69}}{2}}\right)-\frac{1+ i\sqrt{3}}{6}\left(\sqrt[3]{\frac{25+3\sqrt{69}}{2}}\right)\\[0.5cm] k_3=-\frac{1}{3}-\frac{1+i\sqrt{3}}{6}\left(\sqrt[3]{\frac{25}{2}-\frac{3\sqrt{69}}{2}}\right)-\frac{1- i\sqrt{3}}{6}\left(\sqrt[3]{\frac{25+3\sqrt{69}}{2}}\right)\\[0.5cm]$$

Then,

$$\boxed{y(x)=C_1\cdot e^{k_1x}+C_2\cdot e^{k_2x}+C_3\cdot e^{k_3x}-homogeneous\,\,\,solution}$$

2 STEP: Solve an Inhomogeneous Equation

We are looking for a solution in the form

$$y(x)=Ax^5+Bx^4+Cx^3+Dx^2+Ex+F$$

Then,

$$y''=20Ax^3+12Bx^2+6Cx+2D\\[0.3cm] y'''=60Ax^2+24Bx+6C$$

Substitute in the original equation

$$60Ax^2+24Bx+6C+20Ax^3+12Bx^2+6Cx+2D=\\[0.3cm] =Ax^5+Bx^4+Cx^3+Dx^2+Ex+F+x^5\longrightarrow\\[0.3cm] (A+1)x^5+Bx^4+(C-20A)x^3+\\[0.3cm] +(D-60A-12B)x^2+(E-24B-6C)x+(F-6C-2D)=0\\[0.3cm] A+1=0\longrightarrow\boxed{A=-1}\\[0.3cm] \boxed{B=0}\\[0.3cm] C-20A=0\longrightarrow\boxed{C=-20}\\[0.3cm] D-60A-12B=0\longrightarrow\boxed{D=-60}\\[0.3cm] E-24B-6C=0\longrightarrow\boxed{E=-120}\\[0.3cm] F--6C-2D=0\longrightarrow\boxed{F=-240}\\[0.3cm]$$

Conclusion,

$$y(x)=-x^5-20x^3-60x^2-120x-240-inhomogeneous\,\,\,solution$$

General conclusion:

$$\boxed{y(x)=C_1e^{k_1x}+C_2e^{k_2x}+C_3e^{k_3x}-x^5-20x^3-60x^2-120x-240}$$

ANSWER

$$y(x)=C_1e^{k_1x}+C_2e^{k_2x}+C_3e^{k_3x}-x^5-20x^3-60x^2-120x-240$$

Equation #2:

$$y\cdot y''=y'+8x^3$$

1 STEP: Solve a homogeneous equation

$$y\cdot y''=y'\longrightarrow y''=\frac{y'}{y}\longrightarrow\\[0.3cm] \frac{d}{dx}\left(y'\right)=\frac{d}{dx}\left(\ln(y)\right)\longrightarrow y'=\ln(Cy)\longrightarrow\\[0.3cm] \frac{dy}{dx}=\ln(Cy)\longrightarrow\int\frac{dy}{\ln(Cy)}=\int1dx$$

Since

$$\int\frac{dy}{\ln(Cy)} \,is\, the\,integral\,logarithm, not\,an\,elementary\,function$$

This means that we can’t write down the general solution of this equation in the usual elementary functions. The only thing we CAN write is

$$\boxed{li(y)=x+C}$$

More information: https://en.wikipedia.org/wiki/Logarithmic_integral_function