∫ y=e^x(acosx+bsinx) eching

$$\int y = e ^ { x } ( a \cos x + b \sin x )$$

a uchun yechish

$$\left\{\begin{matrix}a=\frac{-b\sin(x)e^{x}+xy+С}{\cos(x)e^{x}}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+\frac{\pi }{2}\\a\in \mathrm{R}\text{, }&С=b\sin(x)e^{x}-xy\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi \left(2n_{1}+1\right)}{2}\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

e^{x} ga a\cos(x)+b\sin(x) ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\int y\mathrm{d}x=e^{x}a\cos(x)+e^{x}b\sin(x)$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$e^{x}a\cos(x)+e^{x}b\sin(x)=\int y\mathrm{d}x$$

Ikkala tarafdan e^{x}b\sin(x) ni ayirish.

$$e^{x}a\cos(x)=\int y\mathrm{d}x-e^{x}b\sin(x)$$

Tenglama standart shaklda.

$$\cos(x)e^{x}a=-b\sin(x)e^{x}+xy+С$$

Ikki tarafini e^{x}\cos(x) ga bo‘ling.

$$\frac{\cos(x)e^{x}a}{\cos(x)e^{x}}=\frac{-b\sin(x)e^{x}+xy+С}{\cos(x)e^{x}}$$

e^{x}\cos(x) ga bo'lish e^{x}\cos(x) ga ko'paytirishni bekor qiladi.

$$a=\frac{-b\sin(x)e^{x}+xy+С}{\cos(x)e^{x}}$$

yx+С-e^{x}b\sin(x) ni e^{x}\cos(x) ga bo'lish.

$$a=\frac{\frac{xy+С}{e^{x}}-b\sin(x)}{\cos(x)}$$

b uchun yechish

$$\left\{\begin{matrix}b=\frac{-a\cos(x)e^{x}+xy+С}{\sin(x)e^{x}}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}\\b\in \mathrm{R}\text{, }&С=a\cos(x)e^{x}-xy\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

e^{x} ga a\cos(x)+b\sin(x) ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\int y\mathrm{d}x=e^{x}a\cos(x)+e^{x}b\sin(x)$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$e^{x}a\cos(x)+e^{x}b\sin(x)=\int y\mathrm{d}x$$

Ikkala tarafdan e^{x}a\cos(x) ni ayirish.

$$e^{x}b\sin(x)=\int y\mathrm{d}x-e^{x}a\cos(x)$$

Tenglama standart shaklda.

$$\sin(x)e^{x}b=-a\cos(x)e^{x}+xy+С$$

Ikki tarafini e^{x}\sin(x) ga bo‘ling.

$$\frac{\sin(x)e^{x}b}{\sin(x)e^{x}}=\frac{-a\cos(x)e^{x}+xy+С}{\sin(x)e^{x}}$$

e^{x}\sin(x) ga bo'lish e^{x}\sin(x) ga ko'paytirishni bekor qiladi.

$$b=\frac{-a\cos(x)e^{x}+xy+С}{\sin(x)e^{x}}$$

yx+С-e^{x}a\cos(x) ni e^{x}\sin(x) ga bo'lish.

$$b=\frac{\frac{xy+С}{e^{x}}-a\cos(x)}{\sin(x)}$$