CAIE A-Level Further Math A2 2.6.6 Initial Conditions QuestionsPractise finding particular differential-equation solutions from initial conditions using Further Mathematics Paper 2 questions and mark schemes.Syllabus2028–2030CourseFurther Mathematics 9231LevelA2
CAIE A-Level Further Math A2 2.6.6 Initial Conditions Questions question 1[Maximum number: 7]Hence find the solution of the differential equation(x2+1)dy dx+yx2+1=x2−xx2+1\left(x^{2}+1\right) \frac{\mathrm{d} y}{\mathrm{~d} x}+y \sqrt{x^{2}+1}=x^{2}-x \sqrt{x^{2}+1}(x2+1) dxdy+yx2+1=x2−xx2+1for which y=ln2y=\ln 2y=ln2 when x=0. Give your answer in the form y=f(x).Mark as masteredShow Answerddx(y(x+(x2+1)))=−xx2+1\frac{\mathrm{d}}{\mathrm{d} x}\left(y\left(x+\sqrt{ }\left(x^{2}+1\right)\right)\right)=-\frac{x}{x^{2}+1}dxd(y(x+(x2+1)))=−x2+1xy(x+(x2+1))=−∫xx2+1 dx=−12ln(x2+1)+Cy\left(x+\sqrt{ }\left(x^{2}+1\right)\right)=-\int \frac{x}{x^{2}+1} \mathrm{~d} x=-\frac{1}{2} \ln \left(x^{2}+1\right)+Cy(x+(x2+1))=−∫x2+1x dx=−21ln(x2+1)+Cln2=C\ln 2=Cln2=Cy=ln2−12ln(x2+1)x+(x2+1)=(x−(x2+1))ln(12(x2+1))y=\frac{\ln 2-\frac{1}{2} \ln \left(x^{2}+1\right)}{x+\sqrt{ }\left(x^{2}+1\right)}=\left(x-\sqrt{ }\left(x^{2}+1\right)\right) \ln \left(\frac{1}{2} \sqrt{ }\left(x^{2}+1\right)\right)y=x+(x2+1)ln2−21ln(x2+1)=(x−(x2+1))ln(21(x2+1))7b 7Add to Test