What is the solution for y^'-5y=2x^2-5x+8 ?

The solution for y^'-5y=2x^2-5x+8 is y=-(2x^2)/5+(21x)/(25)-179/125+c_{1}e^{5x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'-5y=2x^2-5x+8

y^{^{'}} - 5 y = 2 x^{2} - 5 x + 8

y = - \frac{2 x ^{2}}{5} + \frac{21 x}{25} - \frac{179}{125} + c_{1} e^{5 x}

Solution steps

y^{'} (x) - 5 y = 2 x^{2} - 5 x + 8

Solve linear ODE:

y = - \frac{2 x ^{2}}{5} + \frac{21 x}{25} - \frac{179}{125} + c_{1} e^{5 x}

y = - \frac{2 x ^{2}}{5} + \frac{21 x}{25} - \frac{179}{125} + c_{1} e^{5 x}

\frac{d}{d x} (- x^{3} + 20 x^{2} + 21 x)

y^{^{''}} + 4 y = x^{2}

y^{^{'}} = \frac{3 x}{y}

\frac{d}{d x} (\frac{ln ( 3 x ^{2} - 6 x )}{4 x ^{3} - 2})

\int (ln (x))^{7} d x

What is the solution for y^'-5y=2x^2-5x+8 ?
The solution for y^'-5y=2x^2-5x+8 is y=-(2x^2)/5+(21x)/(25)-179/125+c_{1}e^{5x}