Question

The complementary function of $({D^2} + 1)y = {e^{2x}}$ is:
$A.(Ax + B){e^x} \\\ B.A\cos x + B\sin x \\\ C.(Ax + B){e^{2x}} \\\ D.(Ax + B){e^{ - x}} \\\$

Answer 1

Hint: Use auxiliary equation concept and ${y_c} = {e^{ax}}(A\cos \beta x + B\sin \beta x)$to find the complementary function of $({D^2} + 1)y = {e^{2x}}$.
Auxiliary equation is an equation with one variable and equated to zero, which is derived from a given linear differential equation and in which the coefficient and power of the variable in each term correspond to the coefficient and order of a derivative in the original equation.

Complete step-by-step answer:
Hence, the auxiliary equation of above differential equation is $({D^2} + 1)y = 0$
For complementary function let $D = m$
Hence, $\Rightarrow ({m^2} + 1)y = 0$
OR
$\Rightarrow {m^2} + 1 = 0 \\\ \Rightarrow {m^2} = - 1 \\\$
$\Rightarrow$ $m = \sqrt { - 1} = \pm i$
Since the roots are complex , so by formula ${y_c} = {e^{ax}}(A\cos \beta x + B\sin \beta x)$
Where , $a = 0$and $\beta = 1$
Hence , by substituting the values in the formula we get
${y_c} = (A\cos x + B\cos x)$

Note: It is advisable to remember various characteristic roots and their formulas to save time. Eventually it will be difficult to mug up every formula but with practice things get easier.