Understanding Engineering Mathematics

always work, but is a good start, and we will come back to exceptional cases later. This approach is called the method of undete ...

soM=0. Thus, the PI isyp=x, as you can (should!) check by substituting back into the equation: y′′+ 2 y′+y= 0 + 2 ( 1 )+x=x+ 2 T ...

For the PI you might be tempted to tryyp=Lcos 3x, but in fact we need both cos 3x and sin 3xterms because of the first order der ...

Or, cancellinge−xand equating coefficients of cosxand sinx 4 L− 2 M= 1 2 L+ 4 M= 0 These give L= 1 5 ,M=− 1 10 So finally, the G ...

We have yp′′−yp′− 2 yp=L(x− 2 )e−x−L( 1 −x)e−x− 2 Lxe−x =− 3 Le−x=the RHS=e−x SoLxe−xis indeed a solution of the DE ifL=−^13 and ...

(b) If the complementary function contains termsekx,xekx,..., xmekx,takeyp(x)to be of the form yp=Lxm+^1 ekx (4) f(x)of the form ...

Answers (i) 1 4 (ex−e−^3 x) (ii) − 1 4 cos 2x− π 16 sin 2x+ 1 4 x+ 1 4 (iii) 2 3 sinx− 1 3 sin 2x 1 6 e^3 x− 3 2 ex+x+ 4 3 ...

5.Solve the following initial value problems (i) y′− 3 y=e^5 x y( 0 )= 0 (ii) xy′− 2 y=x^2 y( 1 )= 2 (iii) xy′+ 2 y=x^2 y( 1 )= ...

whereλ>0 is some experimentally determined constant of proportionality, andT 0 is the initial temperature. Solve this to give ...

of the body, water flow between connected reservoirs, or financial transactions in a commercial environment. This exercise illus ...

Show that the solution in this case is x=Asin(ωt+φ) whereω^2 =α/m. In the presence of damping, but no forcing terms, the equatio ...

15.9 Answers to reinforcement exercises m=moe−λt (i) x^2 2 + 1 (ii) sinx (iii) 1 2 (x^21 )+lnx (iv) 2x^2 + 2 x+1(v) x^4 12 − x ...

(b) (i) Aex+Be−^2 x− 1 2 x− 3 4 (ii) (Ax+B)e^2 x+ 1 4 x+ 1 2 (iii) e−^2 x(Acosx+Bsinx)+ 1 5 x+ 1 25 (iv) Acos 2x+Bsin 2x+ 1 4 x+ ...

(iv) − 1 2 cos 2x+ 1 2 ( cos 2− 1 sin 2 ) sin 2x+ 1 2 (v) 2 9 [ e^3 − 1 e^6 − 1 ] e^3 x+ 2 9 e^3 [ e^3 − 1 e^6 − 1 ] e−^3 x− 2 9 ...

16 Functions of More than One Variable – Partial Differentiation This chapter deals with functions of more than one variable and ...

16.1 Introduction Often, topics in engineering mathematics can be presented most clearly by means of diagrams rather than symbol ...

(x,y) z x x = f(x,y) y Figure 16.1Surface defined byz=f(x,y). x y z ax+by +c =^0 z = ax +c z = by+ c Figure 16.2The linear funct ...

z z = f(x,y) x (^0) y Tangent plane Figure 16.3The tangent plane. of the slope and rate of change of the surface at the point of ...

16.3 Partial differentiation The lines in the tangent plane parallel to thexz-plane, give the slope of the surface in the x-dire ...

z 0 x (x 0 ,y 0 ,z 0 ) Plane y = y 0 z = f(x,y) y Tangent line with slope∂z ∂x Figure 16.5Definition of the partial derivative ∂ ...