What is the solution for (t^2+1)y^'=sin(t^2+3cos(t))y ?

The solution for (t^2+1)y^'=sin(t^2+3cos(t))y is y=e^{integral of (sin(t^2+3cos(t)))/(t^2+1)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(t^2+1)y^'=sin(t^2+3cos(t))y

(t^{2} + 1) y^{^{'}} = sin (t^{2} + 3 cos (t)) y

y = e^{\int \frac{s i n ( t ^{2} + 3 c o s ( t ) )}{t ^{2} + 1} d t}

Solution steps

(t^{2} + 1) y^{'} (t) = sin (t^{2} + 3 cos (t)) y

Solve separable ODE:

y = e^{\int \frac{s i n ( t ^{2} + 3 c o s ( t ) )}{t ^{2} + 1} d t}

y = e^{\int \frac{s i n ( t ^{2} + 3 c o s ( t ) )}{t ^{2} + 1} d t}

\frac{d y}{d x} = cos^{2} (y), \frac{π}{2} < y

\frac{d y}{d x} (\frac{y}{x}) y = (x + c^{1})^{2}

y^{^{'}} + 2 x \cdot y = 0

2 y (y^{^{'}}) = 6 t^{2}, y (0) = 5

P^{^{'}} = 0.05 P (1 - \frac{P}{400})

What is the solution for (t^2+1)y^'=sin(t^2+3cos(t))y ?
The solution for (t^2+1)y^'=sin(t^2+3cos(t))y is y=e^{integral of (sin(t^2+3cos(t)))/(t^2+1)}