Question

Find the solution to initial value problem d2ydt2−10dydt+25y=0, y(0)=4,y′(0)=5

Answer 1

Consider, pls, this solution.
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Answer 2

The solution to the given initial value problem $\frac{d^2y}{dt^2}-10\frac{dy}{dt} +25y=0$ is $y=-15xe^{5x}+4e^{5x}$

What is differential equation?

"Any equation which contains derivatives is a differential equation."

What is initial value problem?

"It is a differential equation along with an appropriate number of initial conditions."

What is an auxiliary equation?

"An equation obtained from the standard form of a linear differential equation by replacing the right member by zero."

For given question,

We have been given a differential equation $\frac{d^2y}{dt^2}-10\frac{dy}{dt} +25y=0$

And initial conditions $y(0)=4,y'(0)=5$

An auxiliary equation of given differential equation is,

$m^{2} -10m+25=0$

We solve above auxiliary equation.

$\Rightarrow m^{2} -10m+25=0\\\Rightarrow m^{2} -5m-5m+25=0\\\Rightarrow m(m-5)-5(m-5)=0\\\Rightarrow (m-5)(m-5)=0\\\Rightarrow (m-5)^2=0\\\Rightarrow m-5=0\\\Rightarrow m=5$

So, the solution of the given differential equation would be,

$y=C_1xe^{5x}+C_2e^{5x}$

Now we find the value of constants $C_1,C_2$ using the initial values.

Here, y(0) = 4,

$\Rightarrow 4=C_1(0)e^{5\times 0}+C_2e^{5\times 0}\\\Rightarrow 4=0+C_2\times e^0\\\Rightarrow 4=C_2\times 1\\\Rightarrow C_2=4$

For $y=C_1xe^{5x}+C_2e^{5x}$

$y'=C_1e^{5x}+5xC_1e^{5x}+5C_2e^{5x}$

Also, y′(0) = 5

$\Rightarrow 5=C_1e^{5\times 0}+5(0)C_1e^{5\times 0}+5C_2e^{5\times 0}\\\Rightarrow 5=C_1e^0+0+5(4)e^0\\\Rightarrow 5=C_1(1)+20(1)\\\Rightarrow 5=C_1+20\\\Rightarrow C_1=-15$

So, the solution would be $y=-15xe^{5x}+4e^{5x}$

Therefore, the solution to the given initial value problem $\frac{d^2y}{dt^2}-10\frac{dy}{dt} +25y=0$ is $y=-15xe^{5x}+4e^{5x}$

Learn more about the initial value problem here:

https://brainly.com/question/8736446

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