MTH2004 BB Working

FINAL ANSWERS:

1. B
2. ABC
3. AB
4. C

---

$$y^{\prime \prime }+4y=7\sin (2x)$$ $$k^2+4=0:k=\pm 2i⟹\boxed{y_c=c_1\cos (2x)+c_2\sin (2x)}$$ $$\begin{array}{rl}y& =Ax\sin (2x)+Bx\cos (2x)\\ ⟹y^{\prime }& =(B+2Ax)\cos (2x)+(A-2Bx)\sin (2x)\\ ⟹y^{\prime \prime }& =4(A-Bx)\cos (2x)-4(B+Ax)\sin (2x)\end{array}$$ $$(4A)\cos (2x)+(-4B-8Ax)\sin (2x)=(0)\cos (2x)+(7)\sin (2x)$$ $$\{\begin{array}{l}A=0\\ B=-\frac{7}{4}\end{array}⟹\boxed{y_p=-\frac{7}{4}x\cos (2x)}$$ $$\boxed{y=c_1\cos (2x)+c_2\sin (2x)-\frac{7}{4}x\cos (2x)}$$

---

$$y^{\prime \prime }-y^{\prime }-30y=0$$ $$k^2-k-30=0=(k+5)(k-6):k=-5,6⟹\boxed{y_c=c_1{e}^{-5x}+c_2{e}^{6x}}$$ $$W=6e^x+5e^x⟺\boxed{\text{Wronskian}=11e^x}$$

---

$$y^{\prime \prime }+y={\sin }^{2}x$$ $$k^2+1=0:k=\pm i⟹\boxed{y_c=c_1\sin x+c_2\cos x}$$ $$W=-({\sin }^{2}x+{\cos }^{2}x)=-1,f(x)={\sin }^{2}x$$ $$u_1=-\int \frac{y_{2}f(x)}{W}=\int y_{2}f(x)=(\cos x)({\sin }^{2}x)$$ $$u_2=\int \frac{y_{1}f(x)}{W}=-\int y_{1}f(x)=-{\sin }^{3}x$$ $$\boxed{y_p=0}$$

CORRECTION:

$$\begin{array}{rl}{y}_p& =\frac{2}{3}-\frac{1}{3}{\sin }^{2}x\\ & =\frac{2}{3}-\frac{1}{3}(1-{\cos }^{2}x)\\ & =\frac{1}{3}+\frac{1}{3}{\cos }^{2}x\end{array}$$

---

$$x^2{y}^{\prime \prime }+11x{y}^{\prime }+9y=0:\text{Cauchy-Euler}$$ $$m^2+10m+9=0=(m+9)(m+1):m=-9,-1⟹\boxed{y_c=c_1{x}^{-9}+c_2{x}^{-1}}$$ $$W=-x^{-11}+9x^{-11}=8x^{-11},f(x)=0$$ $$\begin{array}{rl}{y}_p& =y_2\int \frac{y_{1}f(x)}{W}-y_1\int \frac{y_{2}f(x)}{W}\\ & =x^{-1}\int \frac{x^{-9}f(x)}{8x^{-11}}-x^{-9}\int \frac{x^{-1}f(x)}{8x^{-11}}\\ & =0\end{array}$$