Solve the differential equation \( \dfrac{dy}{dx} = 3x^2 \) given that \( y = 1 \) when \( x = 0 \).
\( y = x^3 + C \). When \( x=0 \), \( y=1 \) ⇒ \( C=1 \). So \( y = x^3 + 1 \).
Differential Equations
12 exam-style questions on first-order separable equations, integrating factors, second-order linear DEs and modelling. Tap Show answer for full working.
4 Easy4 Medium4 Hard
Solve \( \dfrac{dy}{dx} + 2y = 0 \).
Separable: \( \dfrac{dy}{y} = -2\,dx \) ⇒ \( \ln|y| = -2x + C \) ⇒ \( y = Ae^{-2x} \).
Find the general solution of \( \dfrac{d^2 y}{dx^2} - 9y = 0 \).
Auxiliary: \( m^2 - 9 = 0 \) ⇒ \( m = \pm 3 \). General solution \( y = Ae^{3x} + Be^{-3x} \).
Solve \( \dfrac{dy}{dx} = \dfrac{x}{y} \) given \( y(0) = 2 \).
\( y\,dy = x\,dx \) ⇒ \( \dfrac{y^2}{2} = \dfrac{x^2}{2} + C \). At \( x=0 \), \( y=2 \): \( 2 = C \). So \( y^2 = x^2 + 4 \) (taking positive branch near 0: \( y = \sqrt{x^2+4} \)).
Solve the first-order equation \( \dfrac{dy}{dx} + y = e^{-x} \) using an integrating factor.
IF = \( e^{\int 1\,dx} = e^x \). Then \( \dfrac{d}{dx}(ye^x) = 1 \) ⇒ \( ye^x = x + C \) ⇒ \( y = (x+C)e^{-x} \).
Find the general solution of \( \dfrac{d^2 y}{dx^2} + 4\dfrac{dy}{dx} + 4y = 0 \).
Auxiliary: \( m^2 + 4m + 4 = 0 \) ⇒ \( (m+2)^2 = 0 \), repeated root \( m=-2 \).
\( y = (A + Bx)e^{-2x} \).
Solve \( \dfrac{d^2 y}{dx^2} + y = 0 \) given \( y(0) = 0 \), \( y'(0) = 1 \).
Auxiliary \( m^2 + 1 = 0 \) ⇒ \( m = \pm i \). \( y = A\cos x + B\sin x \).
\( y(0)=0 \) ⇒ \( A=0 \). \( y' = B\cos x \), \( y'(0)=1 \) ⇒ \( B=1 \). So \( y = \sin x \).
Find a particular integral for \( \dfrac{d^2 y}{dx^2} - 3\dfrac{dy}{dx} + 2y = e^{3x} \).
Try \( y_p = ke^{3x} \). Then \( 9k - 9k + 2k = 1 \) ⇒ \( 2k = 1 \) ⇒ \( k = 1/2 \).
Particular integral \( y_p = \dfrac12 e^{3x} \).
Solve \( \dfrac{dy}{dx} + \dfrac{2}{x}y = x \) for \( x > 0 \), given \( y(1) = 2 \).
IF = \( e^{\int 2/x\,dx} = e^{2\ln x} = x^2 \).
\( \dfrac{d}{dx}(yx^2) = x\cdot x^2 = x^3 \) ⇒ \( yx^2 = \dfrac{x^4}{4} + C \) ⇒ \( y = \dfrac{x^2}{4} + \dfrac{C}{x^2} \).
\( y(1)=2 \): \( 1/4 + C = 2 \) ⇒ \( C = 7/4 \). So \( y = \dfrac{x^2}{4} + \dfrac{7}{4x^2} \).
Find the general solution of \( \dfrac{d^2 y}{dx^2} - 2\dfrac{dy}{dx} + 5y = 0 \).
Auxiliary: \( m^2 - 2m + 5 = 0 \) ⇒ \( m = \dfrac{2\pm\sqrt{4-20}}{2} = 1 \pm 2i \).
\( y = e^{x}(A\cos 2x + B\sin 2x) \).
Solve \( \dfrac{d^2 y}{dx^2} + 4y = \sin 2x \), finding both complementary function and a particular integral.
CF: \( m^2 + 4 = 0 \) ⇒ \( m = \pm 2i \), \( y_c = A\cos 2x + B\sin 2x \).
Since \( \sin 2x \) is in CF, try \( y_p = x(Cx\cos 2x + Dx\sin 2x) \)? Or simpler form \( y_p = x(C\cos 2x + D\sin 2x) \).
Standard: \( y_p = x(C\cos 2x + D\sin 2x) \). Differentiating and substituting yields \( C = 0 \), \( D = -1/4 \).
Particular: \( y_p = -\dfrac{x}{4}\sin 2x \). General: \( y = A\cos 2x + B\sin 2x - \dfrac{x}{4}\sin 2x \).
A particle moves so that its displacement \( x \) (metres) from a fixed point at time \( t \) (seconds) satisfies \( \ddot{x} + 4\dot{x} + 13x = 0 \). Given that \( x=3 \) and \( \dot{x}=0 \) when \( t=0 \), find \( x \) in terms of \( t \) and describe the motion.
Auxiliary: \( m^2 + 4m + 13 = 0 \) ⇒ \( m = -2 \pm 3i \).
\( x = e^{-2t}(A\cos 3t + B\sin 3t) \).
\( x(0)=3 \) ⇒ \( A=3 \). \( \dot{x} = -2e^{-2t}(3\cos 3t + B\sin 3t) + e^{-2t}(-9\sin 3t + 3B\cos 3t) \).
\( \dot{x}(0)=0 \): \( -2\cdot 3 + 3B = 0 \) ⇒ \( -6 + 3B = 0 \) ⇒ \( B=2 \).
\( x = e^{-2t}(3\cos 3t + 2\sin 3t) \). Damped oscillatory motion (decaying amplitude).