Question

find the value of y(1) by solving the differential equation y dy/dx=6x²+5,y(0)=2​

Answer 1

Step-by-step explanation:

We have,

$y \frac{dy}{dx} = 6 {x}^{2} + 5$

$\implies \: y dy =( 6 {x}^{2} + 5)dx$

$\implies \: \int y dy = \int( 6 {x}^{2} + 5)dx$

$\implies \: \frac{ {y}^{2} }{2} = 6 \frac{{x}^{3}}{3} + 5x + C$

$\implies \: \frac{ {y}^{2} }{2} = 2{x}^{3} + 5x + C$

Also, we have, y(0) = 2

So,

$\implies \: \frac{ {(2)}^{2} }{2} = 2{(0)}^{3} + (0) + C$

$\implies C = 2$

So, required equation

$\implies \: \frac{ {y}^{2} }{2} = 2{x}^{3} + 5x +2$

$\implies \: {y}^{2} = 4{x}^{3} + 10x +4$

Now, at x = 1,

$\implies \: {y}^{2} = 4 {(1)}^{3} + 10(1) + 4$

$\implies \: y = \sqrt{4 {(1)}^{3} + 10(1) + 4}$

$\implies \: y = \sqrt{4 + 10 + 4}$

$\implies \: y = \sqrt{ 18}$

$\implies \: y =3 \sqrt{2}$

So, $y(1) = 3\sqrt{2}$