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Question: How do you find all solutions of the differential equation \(\dfrac{{{d^2}y}}{{d{x^2}}} = 3y\)?...

How do you find all solutions of the differential equation d2ydx2=3y\dfrac{{{d^2}y}}{{d{x^2}}} = 3y?

Explanation

Solution

According to given in the question we have to find all the solutions of the differential equation d2ydx2=3y\dfrac{{{d^2}y}}{{d{x^2}}} = 3y which is as mentioned in the question. So, first of all to determine the solutions of the differential equation as we can see that the given differential equation is a homogeneous differential equation now we have to look at the auxiliary equation which is the quadratic expression with the coefficient of the derivatives.
Now, we have to solve the auxiliary equation which is the quadratic expression with the coefficient of the derivatives which we have already obtained to determine the roots.
Now, to verify we have to again determine the differentiation of the expression which is obtained after the differentiation.

Complete step-by-step solution:
Step 1: First of all to determine the solutions of the differential equation as we can see that the given differential equation is a homogeneous differential equation now we have to look at the auxiliary equation which is the quadratic expression with the coefficient of the derivatives. Hence,
⇒m2+0m−3=0\Rightarrow {m^2} + 0m - 3 = 0
Step 2: Now, we have to solve the auxiliary equation which is the quadratic expression with the coefficient of the derivatives which we have already obtained to determine the roots. Hence,

⇒m2−3=0 ⇒m2=3 ⇒m=±3  \Rightarrow {m^2} - 3 = 0 \\\ \Rightarrow {m^2} = 3 \\\ \Rightarrow m = \pm \sqrt 3 \\\

Hence, the differentiation of the expression is as below:
⇒y=Ae3x+Be−3x\Rightarrow y = A{e^{\sqrt 3 x}} + B{e^{ - \sqrt 3 x}}
Where, A and B are the arbitrary constant.
Step 3: Now, to verify we have to again determine the differentiation of the expression which is obtained after the differentiation. Hence,
⇒y′=A3e3x−B3e−3x\Rightarrow y' = A\sqrt 3 {e^{\sqrt 3 x}} - B\sqrt 3 {e^{ - \sqrt 3 x}}
Now, we have to verify the hence,

⇒y′′=A33e3x+B33e−3x ⇒y′′=3Ae3x+3B−3x  \Rightarrow y'' = A\sqrt 3 \sqrt 3 {e^{\sqrt 3 x}} + B\sqrt 3 \sqrt 3 {e^{ - \sqrt 3 x}} \\\ \Rightarrow y'' = 3A{e^{\sqrt 3 x}} + 3{B^{ - \sqrt 3 x}} \\\

Hence,
y′′−3y=3Ae3x+3Be−3x−3(3Ae3x+3Be−3x)=0y'' - 3y = 3A{e^{\sqrt 3 x}} + 3B{e^{ - \sqrt 3 x}} - 3(3A{e^{\sqrt 3 x}} + 3B{e^{ - \sqrt 3 x}}) = 0
Hence, we have determined all solutions of the differential equation d2ydx2=3y\dfrac{{{d^2}y}}{{d{x^2}}} = 3y which is y=Ae3x+Be−3x.y = A{e^{\sqrt 3 x}} + B{e^{ - \sqrt 3 x}}.

Note: To solve the auxiliary equation which is the quadratic expression with the coefficient of the derivatives which we have already obtained to determine the roots and A and B are the arbitrary constant.
To verify the differentiation it is necessary that we have to determine the value of y′′y'' and then substitute it in y′′−3yy'' - 3y.