What is (t^2+5t)x^'=5x+5,x(1)=6 ?

The solution to (t^2+5t)x^'=5x+5,x(1)=6 is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(t^2+5t)x^'=5x+5,x(1)=6

(t^{2} + 5 t) x^{^{'}} = 5 x + 5, x (1) = 6

(t = \frac{- 5 \cdot 6 ^{'} + 25 \cdot 3 6 ^{'} + 140 \cdot 6 ^{'}}{2 \cdot 6 ^{'}}, t = \frac{- 5 \cdot 6 ^{'} - 25 \cdot 3 6 ^{'} + 140 \cdot 6 ^{'}}{2 \cdot 6 ^{'}}, x = 6 x = 6)

Solution steps

[(t^{2} + 5 t) x^{^{'}} = 5 x + 5 x (1) = 6]

Solve

x (1) = 6 : x = 6

Plug the solutions

x = 6

into

(t^{2} + 5 t) x^{^{'}} = 5 x + 5

For

(t^{2} + 5 t) x^{^{'}} = 5 x + 5

, subsitute

x

with

6 : t = \frac{- 5 \cdot 6 ^{^{'}} + 25 \cdot 3 6 ^{^{'}} + 140 \cdot 6 ^{^{'}}}{2 \cdot 6 ^{^{'}}}, t = \frac{- 5 \cdot 6 ^{^{'}} - 25 \cdot 3 6 ^{^{'}} + 140 \cdot 6 ^{^{'}}}{2 \cdot 6 ^{^{'}}}

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

(t^{2} + 5 t) x^{^{'}} = 5 x + 5, x (1) = 6

are

t = \frac{- 5 \cdot 6 ^{^{^{'}}} + 25 \cdot 3 6 ^{^{^{'}}} + 140 \cdot 6 ^{^{^{'}}}}{2 \cdot 6 ^{^{^{'}}}}, t = \frac{- 5 \cdot 6 ^{^{^{'}}} - 25 \cdot 3 6 ^{^{^{'}}} + 140 \cdot 6 ^{^{^{'}}}}{2 \cdot 6 ^{^{^{'}}}}, x = 6 x = 6

y^{^{'}} = \frac{1}{5} y cos^{2} (y)

y^{^{'}} + 2 y - 4 = 0

\frac{d y}{d x} + ln (x) \cdot y = 3 \cdot ln (x)

\frac{d x}{d t} = 14 - (\frac{84}{100}) \cdot x

x + y x d x - d y = 0