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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

(1 + x^{2}) y^{'} (x) + 4 x y = x

Solve separable ODE:

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

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

\int \frac{2}{( 1 - 2 x ) ^{2}} d x

\int_{- 1}^{0} e^{- x} d x

\frac{d}{d x} (5 tan (3 x))

x \to 0 - lim (x)

\frac{d y}{d x} = - y - 3

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