What is the solution for y^'+(3x)/(x^2+4)y= x/(x^2+4),y(0)=1 ?

The solution for y^'+(3x)/(x^2+4)y= x/(x^2+4),y(0)=1 is y= 1/3+(16)/(3(x^2+4)^{3/2)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+(3x)/(x^2+4)y= x/(x^2+4),y(0)=1

y^{^{'}} + \frac{3 x}{x ^{2} + 4} y = \frac{x}{x ^{2} + 4}, y (0) = 1

y = \frac{1}{3} + \frac{16}{3 ( x ^{2} + 4 ) ^{\frac{3}{2}}}

Solution steps

y^{'} (x) + \frac{3 x}{x ^{2} + 4} y = \frac{x}{x ^{2} + 4}

Solve linear ODE:

y = \frac{1}{3} + \frac{16}{3 ( x ^{2} + 4 ) ^{\frac{3}{2}}}

y = \frac{1}{3} + \frac{16}{3 ( x ^{2} + 4 ) ^{\frac{3}{2}}}

y = 4 x^{2}, y = 7 x - 3 x^{2}

\frac{2 x}{2 + x}

\int sin^{2} (8 x) d x

\frac{d}{d x} (5 x^{4} - 60 x^{2} + 64)

\int e^{x} x^{- 1} d x

What is the solution for y^'+(3x)/(x^2+4)y= x/(x^2+4),y(0)=1 ?
The solution for y^'+(3x)/(x^2+4)y= x/(x^2+4),y(0)=1 is y= 1/3+(16)/(3(x^2+4)^{3/2)}