Understanding Engineering Mathematics

all possible solutions of the DE – this is then referred to as thegeneral solution.Such solutions contain one or more arbitrary ...

giving the required DE d^2 y dx^2 + 4 y= 0 (b) Similarly: d^2 y dx^2 =−12 cos 2x=− 4 y so d^2 y dx^2 + 4 y= 0 i.e. this function ...

From dy dx =x^2 we have y= ∫ x^2 dx+C= x^3 3 +C which is the required general solution. (a) Ify=1whenx=0thenwehave: y= 1 = 0 +C ...

15.3 First order equations – direct integration and separation of variables We will only consider cases where we can solve the g ...

Problem 15.6 Find the general solution of the DEs: (i) dy dx =xcosx (ii) dy dx =cos^2 y In (ii) find the particular solution sat ...

Note that because of the multi-valued nature of the inverse tan function we have to restrict the region on which the DE is defin ...

STOP AND THINK! The next step is one where many beginners slip up. To solve forywe ‘take the expo- nential’ of both sides (131 ➤ ...

or sincee^0 =1,C=−^32 and the solution can be written e−y= 3 2 − e^2 x 2 Note that sincee−ymust be positive, we are here restric ...

Problem 15.8 By substitutingy=xvfind the general solution of the DE dy dx = xYy x We note that dy dx = 1 + y x =a function of y ...

15.4 Linear equations and integrating factors This class of DEs is absolutely fundamental. Such equations occur throughout scien ...

This can now be solved by direct integration: Iy= ∫ IQdx+C and we finally get the solution y= 1 I (∫ IQdx+C ) = 1 I ∫ IQdx+ C I ...

Dividing byxto make the coefficient of the derivative unity gives dy dx + 1 x y=x Problem 15.10 Write down the DE satisfied by t ...

(back where we started, but bear with me on this – we’ll come back to it later). We know that the left-hand side must now be the ...

Answer (i) y= x 2 + C x (ii) y=x^3 +Cx^2 − 1 15.5 Second order linear homogeneous differential equations If you found the previo ...

Any particular solution of theinhomogeneousequation is aparticular integralof the equation. By substituting it in the equation y ...

Sinceeλx =0wehave aλ^2 +bλ+c= 0 Soλsatisfies a quadratic equation with the same coefficients as the DE itself,ay′′+by′+ cy. T ...

or with C=A+BandD=A−jB y=eαx(Ccosβx+Dsinβx) Note: Although we usually prefer such a real form of the solution, particularly in a ...

Solving these two equations givesA=1,B=−1 and the particular solution satisfying the initial conditions is y=e−x−e−^2 x Problem ...

The AE is λ^2 + 2 λ+ 2 = 0 which has the complex roots λ=− 1 ±j This gives the two solutions e(−^1 +j)x,e(−^1 −j)x and the gener ...

x x y y a = 0 a < 0 Figure 15.2Undamped and damped oscillations. xincreases. This could represent an oscillating system in a ...