What is the solution for (dy)/(dt)= t/(t^2y+y),y(0)=7 ?

The solution for (dy)/(dt)= t/(t^2y+y),y(0)=7 is y=sqrt(ln(t^2+1)+49)

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(dy)/(dt)= t/(t^2y+y),y(0)=7

\frac{d y}{d t} = \frac{t}{t ^{2} y + y}, y (0) = 7

y = ln (t^{2} + 1) + 49

Solution steps

\frac{d y}{d t} = \frac{t}{t ^{2} y + y}

Solve separable ODE:

y = ln (t^{2} + 1) + 49

y = ln (t^{2} + 1) + 49

x \to 7 - lim (∣ x - 7 ∣)

n = 1 \sum \infty \frac{7}{n 8 ^{n}}

\int_{- \infty}^{- 1} \frac{3}{x ^{2}} d x

y^{^{''}} + 8 y^{^{'}} + 32 y = - 320 sin (4 x)

\frac{d y}{d x} + \frac{y}{x ( x - 1 )} = (x - 1)

What is the solution for (dy)/(dt)= t/(t^2y+y),y(0)=7 ?
The solution for (dy)/(dt)= t/(t^2y+y),y(0)=7 is y=sqrt(ln(t^2+1)+49)