Question

If we have a differential equation ${y}''+9y=2{{x}^{2}}-5$ , How about the solution. $\left( {{y}_{h}}=?,{{y}_{p}}=? \right)$

Answer 1

Hint: The above-given equation is the nonhomogeneous differential equation of second order. The right approach to this problem is by writing an auxiliary equation of the above-given function. We then find the values of ${{y}_{h}}$ and ${{y}_{p}}$to find the general solution of the function.

Complete step-by-step solution:
The given equation in the above question is given by,
  $\Rightarrow {y}''+9y=2{{x}^{2}}-5$
The homogeneous equation for the above-written equation is,
$\Rightarrow {y}''+9y=0$
We need to write an auxiliary equation for the above equation,
An auxiliary equation is an equation that is written for all linear differential equations.
The auxiliary equation is the polynomial equation with coefficient as derivatives.
$\Rightarrow {{m}^{2}}+9=0$
$\Rightarrow {{m}^{2}}=-9$
$\Rightarrow m=\pm (-3)$
The values of m are purely imaginary which means that the roots of the given auxiliary equation are imaginary.
Roots of the auxiliary equation and its solutions:
1. If the roots of the auxiliary equation are real and repeated then the solution is of the form $y=\left( Ax+B \right){{e}^{\alpha x}}$ for the repeated root $m=\alpha$.
2. If the roots of the auxiliary equation are real and distinct such as $m={{\alpha }_{1}},{{\alpha }_{2}}....$ then the solution is of the form ${{y}_{1}}=A{{e}^{{{\alpha }_{1}}x}},{{y}_{2}}=B{{e}^{{{\alpha }_{2}}x}},....$
3. If the roots of the auxiliary equation are imaginary and are in the form of $m=a\pm bi$ then the solution is of the form $y={{e}^{ax}}\left( A\cos \left( bx \right)+B\sin \left( bx \right) \right)$
From the above,
The solution of the homogeneous equation ${y}''+9y=0$ is
$\Rightarrow y={{e}^{0x}}\left( A\cos \left( 3x \right)+B\sin \left( 3x \right) \right)$
$\Rightarrow y=\left( A\cos \left( 3x \right)+B\sin \left( 3x \right) \right)$
Therefore, ${{y}_{h}}=\left( A\cos \left( 3x \right)+B\sin \left( 3x \right) \right)$
Particular solution:
We find the solution of the non-homogeneous equation $2{{x}^{2}}-5$
The above equation is of the form $y=a{{x}^{2}}+bx+c$
Differentiating the equation y,
$\Rightarrow {y}'=2ax+b$
Differentiating the above equation,
$\Rightarrow {y}''=2a$
Substituting the values of $y,{y}''$in the equation ${y}''+9y=2{{x}^{2}}-5$, we get,
$\Rightarrow \left( 2a \right)+9(a{{x}^{2}}+bx+c)=2{{x}^{2}}-5$
Comparing the coefficients of ${{x}^{2}},x,{{x}^{0}}$ on both sides, we get,
Coefficients of ${{x}^{2}}$:
$\Rightarrow 9a=2$
$\Rightarrow a=\dfrac{2}{9}$
Coefficients of $x$:
$\Rightarrow 9b=0$
$\Rightarrow b=0$
Coefficients of ${{x}^{0}}$:
$\Rightarrow 2a+9c=-5$
Substitute the value of ‘a’ and then evaluate.
$\Rightarrow 2\left( \dfrac{2}{9} \right)+9c=-5$
$\Rightarrow 9c=-5-2\left( \dfrac{2}{9} \right)$
Now evaluate further.
$\Rightarrow 9c=-5-\left( \dfrac{4}{9} \right)$
$\Rightarrow 9c=-\dfrac{49}{9}$
$\Rightarrow c=\dfrac{-49}{81}$
Therefore, ${{y}_{p}}=\dfrac{2}{9}{{x}^{2}}-\dfrac{49}{81}$
The general solution is ${{y}_{h}}+{{y}_{p}}$
Hence, the general solution is $A\cos 3x+B\sin 3x+\dfrac{2}{9}{{x}^{2}}-\dfrac{49}{81}$

Note: The general solution of the function is the sum of the complementary function and the particular solution. We should remember the solutions of auxiliary equations to solve the problem easily.