Pearson Edexcel IAL Mathematics FP2.4.3 Reducible first order differential equationsPractise using given substitutions to reduce first order differential equations, then solve the resulting linear equation.SyllabusFirst assessment 2019CourseMathematics YMA01LevelA2
Exam pointsshow the transformed linear equation before solving it by an integrating factorconvert back to y and use an initial condition to give the particular solution
FP2.4.3 - Reducible first order differential equations question 1[Maximum number: 5]Question (a)(a)Show that the transformation v=y-2 x transforms the differential equationdy dx+2yx(y−4x)=2−8x3\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y x(y-4 x)=2-8 x^{3} dxdy+2yx(y−4x)=2−8x3into the differential equationdv dx=−2xv2\frac{\mathrm{d} v}{\mathrm{~d} x}=-2 x v^{2} dxdv=−2xv2[ 4 ]Show AnswerMark as mastereddv dx=dy dx−2\frac{\mathrm{d} v}{\mathrm{~d} x}=\frac{\mathrm{d} y}{\mathrm{~d} x}-2 dxdv= dxdy−2B1dy dx+2yx(y−4x)=2−8x3→ dv dx+2+2(v+2x)x(v+2x−4x)=2−8x3\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y x(y-4 x)=2-8 x^{3} \rightarrow \frac{\mathrm{~d} v}{\mathrm{~d} x}+2+2(v+2 x) x(v+2 x-4 x)=2-8 x^{3} dxdy+2yx(y−4x)=2−8x3→ dx dv+2+2(v+2x)x(v+2x−4x)=2−8x3}→dv dx+2+2x(v2−4x2)=2−8x3\rightarrow \frac{\mathrm{d} v}{\mathrm{~d} x}+2+2 x\left(v^{2}-4 x^{2}\right)=2-8 x^{3}→ dxdv+2+2x(v2−4x2)=2−8x3М1→dv dx=−2xv2∗\rightarrow \frac{\mathrm{d} v}{\mathrm{~d} x}=-2 x v^{2} *→ dxdv=−2xv2∗A1*(4)Question (b)(b)Hence obtain the general solution of the differential equation (I).[ 1 ]Show Answery=2x+1x2+cy=2 x+\frac{1}{x^{2}+c}y=2x+x2+c1B1ft(1)Add to Test
Question (a)(a)Show that the transformation v=y-2 x transforms the differential equationdy dx+2yx(y−4x)=2−8x3\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y x(y-4 x)=2-8 x^{3} dxdy+2yx(y−4x)=2−8x3into the differential equationdv dx=−2xv2\frac{\mathrm{d} v}{\mathrm{~d} x}=-2 x v^{2} dxdv=−2xv2[ 4 ]Show AnswerMark as mastereddv dx=dy dx−2\frac{\mathrm{d} v}{\mathrm{~d} x}=\frac{\mathrm{d} y}{\mathrm{~d} x}-2 dxdv= dxdy−2B1dy dx+2yx(y−4x)=2−8x3→ dv dx+2+2(v+2x)x(v+2x−4x)=2−8x3\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y x(y-4 x)=2-8 x^{3} \rightarrow \frac{\mathrm{~d} v}{\mathrm{~d} x}+2+2(v+2 x) x(v+2 x-4 x)=2-8 x^{3} dxdy+2yx(y−4x)=2−8x3→ dx dv+2+2(v+2x)x(v+2x−4x)=2−8x3}→dv dx+2+2x(v2−4x2)=2−8x3\rightarrow \frac{\mathrm{d} v}{\mathrm{~d} x}+2+2 x\left(v^{2}-4 x^{2}\right)=2-8 x^{3}→ dxdv+2+2x(v2−4x2)=2−8x3М1→dv dx=−2xv2∗\rightarrow \frac{\mathrm{d} v}{\mathrm{~d} x}=-2 x v^{2} *→ dxdv=−2xv2∗A1*(4)
Question (b)(b)Hence obtain the general solution of the differential equation (I).[ 1 ]Show Answery=2x+1x2+cy=2 x+\frac{1}{x^{2}+c}y=2x+x2+c1B1ft(1)