Question

Solve the initial value problems:
a) y' = (x y sin(x))/(y+1), y(0) = 1
b) t(du/dt) = t^2 + 3u, t>0, u(2)=4

Answer 1

Answer on Question #54160 - Math - Calculus

Solve the initial value problems:

a) $y' = (x y \sin(x)) / (y + 1), y(0) = 1$

b) $t(\mathrm{du} / \mathrm{dt}) = t^2 + 3u, t > 0, u(2) = 4$

Solution

a) $y' = \frac{xysinx}{y + 1}, y(0) = 1.$

$$\begin{array}{l} \frac{y + 1}{y} dy = xsinx dx \rightarrow \int \frac{y + 1}{y} dy = \int xsinx dx \rightarrow \\ \rightarrow \int \left(1 + \frac{1}{y}\right) dy = - \int x \, d\cos(x) \rightarrow \\ \rightarrow y + lny = - (x \cos(x) - \int \cos(x) \, dx) + C \rightarrow \\ \rightarrow y + lny = - x \cos x + sinx + C; \quad (1) \\ y(0) = 1 \quad (2) \end{array}$$

Equalities (1) and (2) give the following expression:

$$\begin{array}{l} y(0) + lny(0) = -0 \cdot \cos(0) + \sin(0) + C \rightarrow \\ 1 = C \quad (3) \end{array}$$

Substitute (3) for $C$ in (1) and obtain the answer.

Solution to the initial value problem is

$$y + lny = - x \cos x + \sin x + 1.$$

b) $t \frac{du}{dt} = t^2 + 3u, t > 0, u(2) = 4.$

Rewrite $t \frac{du}{dt} = t^2 + 3u$ as

$$\frac{du}{dt} - \frac{3}{t} u = t$$ $$\frac{du}{dt} + P(t)u = f(t) \quad \text{where} \quad P(t) = -\frac{3}{t}, \quad f(t) = t$$

Integrating factor is the following:

$$e^{\int P(t)dt} = e^{-3lnt} = t^{-3}$$

Multiply (4) by (5):

$$\frac{1}{t^3} \frac{du}{dt} - \frac{3}{t^4} u = \frac{t}{t^3}$$ $$\frac{d}{dt} \left( \frac{u}{t^3} \right) = \frac{1}{t^2}$$

Integrate both sides with respect to $t$ :

$$\frac{u}{t^3} = \int \frac{dt}{t^2} \rightarrow \frac{u}{t^3} = -\frac{1}{t} + C \rightarrow$$

Multiply both sides by $t^3$ :

$$u = C t^3 - t^2;$$

It follows from the initial condition $u(2) = 4$ and (6) that

$$u(2) = 4 \rightarrow 4 = 8C - 4 \rightarrow$$ $$C = 1$$

Substitute (7) for $C$ in (6) and obtain the answer.

Solution to the initial value problem is

$$\boldsymbol{u} = \boldsymbol{t}^3 - \boldsymbol{t}^2.$$

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