CAIE A-Level Further Math A2 2.6.4 Linear Differential Equations QuestionsPractise finding general solutions of higher-order linear differential equations with Further Mathematics Paper 2 questions and mark schemes.Syllabus2028–2030CourseFurther Mathematics 9231LevelA2
CAIE A-Level Further Math A2 2.6.4 Linear Differential Equations Questions question 1[Maximum number: 7]It is given that x=t3yx=t^{3} yx=t3y andt3 d2y dt2+(4t3+6t2)dy dt+(13t3+12t2+6t)y=61e12t.t^{3} \frac{\mathrm{~d}^{2} y}{\mathrm{~d} t^{2}}+\left(4 t^{3}+6 t^{2}\right) \frac{\mathrm{d} y}{\mathrm{~d} t}+\left(13 t^{3}+12 t^{2}+6 t\right) y=61 \mathrm{e}^{\frac{1}{2} t} .t3 dt2 d2y+(4t3+6t2) dtdy+(13t3+12t2+6t)y=61e21t.Find the general solution for y in terms of t.Show Answerm2+4m+13=0⇒m=−2±3im^{2}+4 m+13=0 \Rightarrow m=-2 \pm 3 im2+4m+13=0⇒m=−2±3ix=e−2t(Acos3t+Bsin3t)x=\mathrm{e}^{-2 t}(A \cos 3 t+B \sin 3 t)x=e−2t(Acos3t+Bsin3t)x=ke12t⇒x˙=12ke12t⇒x¨=14ke12tx=k \mathrm{e}^{\frac{1}{2} t} \Rightarrow \dot{x}=\frac{1}{2} k \mathrm{e}^{\frac{1}{2} t} \Rightarrow \ddot{x}=\frac{1}{4} k \mathrm{e}^{\frac{1}{2} t}x=ke21t⇒x˙=21ke21t⇒x¨=41ke21t14k+2k+13k=61⇒k=4\frac{1}{4} k+2 k+13 k=61 \Rightarrow k=441k+2k+13k=61⇒k=4t3y=e−2t(Acos3t+Bsin3t)+4e12t⇒y=t−3e−2t(Acos3t+Bsin3t)+4t−3e12tt^{3} y=\mathrm{e}^{-2 t}(A \cos 3 t+B \sin 3 t)+4 \mathrm{e}^{\frac{1}{2} t} \Rightarrow y=t^{-3} \mathrm{e}^{-2 t}(A \cos 3 t+B \sin 3 t)+4 t^{-3} \mathrm{e}^{\frac{1}{2} t}t3y=e−2t(Acos3t+Bsin3t)+4e21t⇒y=t−3e−2t(Acos3t+Bsin3t)+4t−3e21t7b 7Add to Test