Question

Q.8. Solve dx/dt=t+4 at x=0,t=0​

Answer 1

Hyy mate❤

Here is your answer

dx/dt = t + 4 at x = 0, t = 0

d×0/d×0 = 0 + 4

0/0 = 4

(0/0 is undefined)

= 4

Hope it helps u...☺

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Answer 2

The required solution is 2x = t² + 8t

Given :

The differential equation

$\displaystyle \sf{ \frac{dx}{dt} = t + 4 }$

To find :

The solution at x = 0 , t = 0

Solution :

Step 1 of 2 :

Write down the given differential equation

The differential equation is

$\displaystyle \sf{ \frac{dx}{dt} = t + 4 }$

Step 2 of 2 :

Solve the differential equation

$\displaystyle \sf{ \frac{dx}{dt} = t + 4 }$

$\displaystyle \sf{ \implies dx = (t + 4)dt}$

On integration we get

$\displaystyle \sf{ \int \: dx = \int \: (t + 4)dt}$

$\displaystyle \sf{ \implies x = \frac{ {t}^{2} }{2} + 4t + \frac{c}{2} }$

$\displaystyle \sf{ \implies 2x = {t}^{2} + 8t + c } \: \: \: - - - (1)$

Where c is constant

Step 3 of 3 :

Find the particular solution

Putting x = 0 , t = 0 in Equation 1 we get

$\displaystyle \sf{ 2 \times 0 = {0}^{2} + 8 \times 0 + c }$

$\displaystyle \sf{ \implies 0 = c }$

$\displaystyle \sf{ \implies c = 0 }$

Hence the required solution is

$\displaystyle \sf{ 2 x = {t}^{2} + 8t }$