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

$$\frac{d y}{d x}+p(x) y=f(x) y^{n}$$

$$\frac{d y}{d x}-\left(\frac{1}{x}+1\right) y=y^{2}$$

$$n=2 \quad u=y^{1-n}=y^{1-2}=y^{-1}$$

$$u=y^{-1}, y=u^{-1}$$

$$\frac{d y}{d x}=-u^{-2} \frac{d u}{d x}$$

$$-u^{-2} \frac{d u}{d x}-\left(\frac{1}{x}+1\right) u^{-1}=u^{-2}$$


$$\frac{d u}{d x}+\left(\frac{1}{x}+1\right) u=-1$$

Integrating factor

$$\frac{d y}{d x}+p(x) y=Q(x)$$

$$I F=e^{\int p(x) d x}$$

$$I F=e^{\int\left(\frac{1}{x}+1\right) d x}=e^{\ln x+x}=e^{x} e^{\ln x}$$

$$I F=x e^{x}$$

$$x e^{x} \frac{d u}{d x}+\left(e^{x}+x e^{x}\right) u=-x e^{x}$$

$$\frac{d}{d x}[I F \cdot u]=-x e^{x}$$

$$\int \frac{d}{d x}[{IF} \cdot u]=\int -x e^{x}$$

IF $$u=\int-x e^{x} d x$$

$$x e^{x} \cdot u=-\left[u v-\int v d u\right]=-\left[x e^{x}-\int e^{x}\right] dx$$

$$x e^{x} \cdot u=-x e^{x}+e^{x}+C$$


$$x e^{x} y^{-1}=-x e^{x}+e^{x}+c \quad \div x e^{x}$$

$$y^{-1}=-1+\frac{1}{x}+\frac{c}{x e^{x}}$$

$$y^{1 / 2} \frac{d y}{d x}+y^{3 / 2}=1 \quad, y(0)=4$$

$$\frac{d y}{d x}+p\left(x\right) y=f(x) y^{n}$$

$$\frac{d y}{d x}+y=y^{-1 / 2} \quad n=-1 / 2$$


$$u=y^{3 / 2}, y=u^{2 / 3}, \frac{d y}{d x}=\frac{2}{3} u^{-1 / 3} \frac{d u}{d x}$$

$$\frac{2}{3} u^{-1 / 3} \frac{d u}{d x}+u^{2 / 3}=\left(u^{2 / 3}\right)^{-1 / 2}=u^{-1 / 3}$$

$$\frac{d u}{d x}+\frac{3}{2} u=\frac{3}{2}$$

$$\frac{d u}{d x}=\frac{3}{2}-\frac{3}{2} u=\frac{3}{2}(1-u)$$

$$\frac{d u}{1-u}=\frac{3}{2} d x$$

$$\frac{-d u}{1-u}=\frac{-3}{2} d x$$

$$\ln |1-u|=\frac{-3}{2} x+C$$

$$1-u=e^{-3 x / 2}+e^{c}$$

$$e^{-3 x/2+c}$$

$$e^{-3 x / 2} \cdot e^{c}$$

$$u=1-e^{-3 x / 2} e^{c}$$

$$u=y^{3 / 2}$$

$$y^{3 / 2}=1-e^{-3 x / 2} e^{c}$$


$$y^{3 / 2}=1+e^{-3 x / 2} c_{1}$$




$$y^{3 / 2}=1+7 e^{-3 x / 2}$$

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