Question

Solve the following Second Order equation with constant co-efficient

y’’ - 6y’+25 y = 0.

Solve the following second order non-homogenous equation:

y" + y = sin (2x)

Solve the following Bernoulli's initial value problem differential equation:

6y’ — 2y=xy⁴ y(0)=–2

Solve the following initial value problem equation:

2x²y" + 3xy' 15y = 0, y (1)=0 y' (1) = 1

Answer 1

$\left(\mathrm{1}\right)\ \ y''{}{}\ -\ \ \mathrm{6}y'\ \ +\ \ \ \mathrm{25}y\ \ =0 \\ A.E.\ \ \ \ m^{\mathrm{2}}-\mathrm{6}m+\mathrm{25}=0 \\ m=\frac{\mathrm{6}\pm \sqrt{\mathrm{36}-\mathrm{100}}}{\mathrm{2}} \\ m=\frac{\mathrm{6}\pm \mathrm{8}i}{\mathrm{2}}\ \ =\ \ \mathrm{3}\pm \mathrm{4}i \\ y=e^{\mathrm{3}x}\left(A\mathrm{cos}\left(\mathrm{4}x\right)\ \ +\ \ B\mathrm{sin}\left(\mathrm{4}x\right)\right) \\ \\ \left(\mathrm{2}\right)\ \ y''{}{}\ +y\ \ =\mathrm{sin}\left(\mathrm{2}x\right) \\ A.E.\ \ \ m^{\mathrm{2}}+\mathrm{1}=0\ \ \\ m=\pm i \\ y_c\ \ =A\mathrm{cos}\left(x\right)+B\mathrm{sin}\left(x\right)......\left(\mathrm{1}\right) \\ Let\ \ y_p=C\ \mathrm{cos}\left(\mathrm{2}x\right)\ \ \ +\ \ D\ \mathrm{sin}\left(\mathrm{2}x\right)\ \ \\ Then\ \ {y''}_p=-\mathrm{4}C\ \mathrm{cos}\left(\mathrm{2}x\right)\ \ \ -\ \ \ \mathrm{4}D\ \mathrm{sin}\left(\mathrm{2}x\right)\ \ \\ \left(-\mathrm{4}C\ \mathrm{cos}\left(\mathrm{2}x\right)\ \ \ -\ \ \ \mathrm{4}D\ \mathrm{sin}\left(\mathrm{2}x\right)\ \ \right)+\left(-\mathrm{4}C\ \mathrm{cos}\left(\mathrm{2}x\right)\ \ \ -\ \ \ \mathrm{4}D\ \mathrm{sin}\left(\mathrm{2}x\right)\ \ \right)\ \ =\ \ \mathrm{sin}\left(\mathrm{2}x\right) \\ collecting\ \ all\ \ common\ \ terms\ and\ \ solve\ \ ,C=0\ \ ,\ \ D=\frac{-\mathrm{1}}{\mathrm{8}} \\ y_p=-\frac{\mathrm{sin}\left(\mathrm{2}x\right)\ }{\mathrm{8}}\ ..............\left(\mathrm{2}\right) y=y_c+y_p=-\frac{\mathrm{sin}\left(\mathrm{2}x\right)\ }{\mathrm{8}}\ \ +A\mathrm{cos}\left(x\right)+B\mathrm{sin}\left(x\right) \\ y=A\mathrm{cos}\left(x\right)+B\mathrm{sin}\left(x\right)-\frac{\mathrm{sin}\left(\mathrm{2}x\right)\ }{\mathrm{8}}\ \\ \\ \mathrm{(3)}\ \ \ \ \mathrm{6}y'\ \ -\mathrm{2}y\ \ =\ xy^{\mathrm{4}}\ ,\ \ \ \ \ y\left(0\right)=-\mathrm{2} \\ y^{-\mathrm{4}}y'\ \ -\frac{y^{-\mathrm{3}}}{\mathrm{3}}\ \ =\ \frac{x}{\mathrm{6}}\ \ \ \ \left(Bernoulli\ \ equation\right) \\ let\ \ z=\ \ \frac{y^{-\mathrm{3}}}{\mathrm{3}}\ \ \ \ ,\ \ z'=-\mathrm{3}y^{-\mathrm{4}}y'\ \ \\ Then\ \ we\ \ have\ \ \mathrm{:}\ \ z'\ +\mathrm{3}z\ \ =\ -\frac{x}{\mathrm{2}}\ \ \ \ \left(Linear\ \ equation\right) \\ z'\ +\mathrm{3}z\ \ =\ -\frac{x}{\mathrm{2}}\ \\ IF\ \ =\ \ e^{\int{\mathrm{3}dx}}\ \ =\ \ e^{\mathrm{3}x} ze^{\mathrm{3}x}\ \ \ =\ \ \ \int{-\frac{xe^{\mathrm{3}x}}{\mathrm{2}}\ }\ \ dx\ \ \ +\ \ \ C \\ \\ ze^{\mathrm{3}x}\ \ =-\frac{xe^{\mathrm{3}x}}{\mathrm{6}}+C \\ \\ z=\frac{-x}{\mathrm{6}}+Ce^{-\mathrm{3}x} \\ \\ \ \ \frac{y^{-\mathrm{3}}}{\mathrm{3}}\ \ \ \ =\frac{-x}{\mathrm{6}}+Ce^{-\mathrm{3}x} \\ \\ but\ \ y\left(0\right)=-\mathrm{2}\ \ \ ,\ \ such\ \ that\ \ \frac{-\mathrm{1}}{\mathrm{24}}=C \\ \frac{y^{-\mathrm{3}}}{\mathrm{3}}\ \ \ \ =\frac{-x}{\mathrm{6}}-\frac{e^{-\mathrm{3}x}}{\mathrm{24}} \\ \\ y^{-\mathrm{3}}\ \ \ \ =-\frac{x}{\mathrm{2}}-\frac{e^{-\mathrm{3}x}}{\mathrm{8}} \\ \\ \left(\mathrm{4}\right)\ \ \mathrm{2}x^{\mathrm{2}}y''\ \ +\ \ \mathrm{3}xy'\ -\ \ \mathrm{15}y\ \ =0\ \ \\ let\ \ y=Ax^r\ \ ,\ \ \ y'=Arx^{r-\mathrm{1}}\ \ ,\ \ y''\ =\ \ Ar\left(r-\mathrm{1}\right)x^{r-\mathrm{2}}\ \\ \mathrm{2}x^{\mathrm{2}}\left(Ar\left(r-\mathrm{1}\right)x^{r-\mathrm{2}}\ \right)\ \ +\ \ \mathrm{3}x\left(Arx^{r-\mathrm{1}}\ \ \right)\ -\mathrm{15}Ax^r\ \ =0 \\ \mathrm{2}r\left(r-\mathrm{1}\right)+\mathrm{3}r-\mathrm{15}=0 \\ \mathrm{2}r^{\mathrm{2}}+r-\mathrm{15}=0 \\ \left(\mathrm{2}r-\mathrm{5}\right)\left(r+\mathrm{3}\right)=0 \\ r=\frac{\mathrm{5}}{\mathrm{2}}\ \ ,\ \ r=-\mathrm{3} \\ y=A_{\mathrm{1}}x^{\frac{\mathrm{5}}{\mathrm{2}}}{}{}{}\ \ +\ \ \ A_{\mathrm{2}}x^{-\mathrm{3}} \\ \\ but\ \ y\left(\mathrm{1}\right)=0\ \ \ \ ,\ \ \ A_{\mathrm{1}}+\ \ \ A_{\mathrm{2}}=0\ \ .....\left(\mathrm{1}\right) \\ and\ \ y\mathrm{'}\left(\mathrm{1}\right)=\mathrm{1}\ \ ,\ \ \frac{\mathrm{5}A_{\mathrm{1}}}{\mathrm{2}}-\mathrm{3}\ A_{\mathrm{2}}=\mathrm{1}\ \ ...\left(\mathrm{2}\right) \\ Solving\ \ \ \mathrm{1}\ \ \ \ and\ \ \ \ \mathrm{2}\ \ \ , \\ A_{\mathrm{1}}=\ \frac{\mathrm{2}}{\mathrm{11}}\ \ \ ,\ \ \ \ A_{\mathrm{2}}=\frac{-\mathrm{2}}{\mathrm{11}}\ \ \ , \\ \\ y=\ \frac{\mathrm{2}}{\mathrm{11}}x^{\frac{\mathrm{5}}{\mathrm{2}}}{}{}{}\ \ -\ \ \ \ \frac{\mathrm{2}}{\mathrm{11}}x^{-\mathrm{3}}$